MAT 300 Unit 5 Challenges Sophia MAT300 Unit 5 Challenges Sophia

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CHALLENEG 1 Which of the following is an example of a statistic? a.)

Average number of all high school graduates who will attend college b.)

Standard deviation of tuition costs for all private universities c.)

Summary value of a population d.)

25 random students are asked how much they spent on books  A sample is a subset of the entire group of interest. A statistic is something that comes from a sample. Since these 25 students are only a subset of all in the population, this is an example of a statistic from a sample.

Which of the following is an example of a parameter? a.)

30 random adults are asked how much they spent on health insurance b.)

50 random students are asked how much they spent on supplies c.)

Average number of all adults who smoke regularly d.)

Summary value of a sample  Recall a parameter comes from the entire set of interest, the population. Since they are looking at all adults here, the average number of adult smokers would be an example of a parameter.

Which of the following is an example of a statistic? a.) $18,000 is the average cost to attend 30 randomly selected public universities b.)

Average number of all high school graduates who will attend college c.)

A value used to represent a population d.)

Standard deviation of all financial aid packages for private universities  A sample is a subset of the entire group of interest. A statistic is something that comes from a sample. Since these 30 universities are only a subset of all in the population, this is an example of a statistic from a sample.

Select two symbols that represent a statistic.

a.) b.) c.) d.) e.) Answer Rationale Recall that p-hat (pp) is the sample proportion and x-bar (xx) is the sample mean. Since all of these come from samples, they are statistics.

Select two symbols that represent a parameter. a.) b.) c.) d. e.) Answer Rationale Recall that sigma (σ) is the population standard deviation and mu (µ) is the population mean.

Since all these values come from the population, they are parameters.

One condition for performing a hypothesis test is that the observations are independent.

If Jason is sampling 35 students without replacement, the minimum population size would have to be greater than ________ to treat the observations as independent. a.)

350 b.)

300 c.)

70 d.)

250  In general, we want the population to be at least 10 times larger than the sample. If the sample is 35, then the minimum population size would be at least 35 * 10 = 350.

One condition for performing a hypothesis test is that the observations are independent.

If Myra is sampling 30 students without replacement, the minimum population size would have to be greater than ________ to treat the observations as independent. a.)

600 b.)

300

c.)

150 d.)

60  In general, we want the population to be at least 10 times larger than the sample. If the sample is 30, then the minimum population size would be at least 30 * 10 = 300.

One condition for performing a hypothesis test is that the observations are independent.

If Blake is sampling 40 students without replacement, the minimum population size would have to be greater than ________ to treat the observations as independent. a.)

600 b.)

80 c.)

200 d.)

400  In general, we want the population to be at least 10 times larger than the sample. If the sample is 40, then the minimum population size would be at least 40 * 10 = 400.

Select the TRUE statement regarding the relationship between sampling error and sample size. a.)

As sample size decreases, sampling error decreases. b.)

As sample size increases, sampling error decreases. c.)

As sample size increases, sampling error increases. d.)

Sample size does not have an impact on sampling error.  If you recall that in the sampling error calculations for mean and proportion, the sample size is in the denominator. So as sample size increases, we can note the sampling error calculation must necessarily decrease. elect the FALSE statement regarding the relationship between sampling error and sample size. a.)

An estimate becomes more accurate as you increase the sample size. b.)

You can resolve poor data collection by increasing the sample size. c.)

As a sample size decreases, the standard deviation increases. d.)

Increasing the sample size will decrease the sampling error.  Recall that sampling process must be randomly drawn. If you don't draw a sample randomly, then increasing sample size will not alter the sampling error.

Select the FALSE statement regarding the relationship between sampling error and sample size. a.)

The sample size can impact the sampling error. b.)

The larger a sample size, the more accurate an estimate can be. c.)

The standard error increases as the sample size increases. d.)

In order to decrease sampling error, you must increase the sample size.  Recall that in general that an increase in the sample size, if randomly drawn, results in a standard error decrease.

If a sample size is greater than 30, which of the following characteristics of the distribution of sample means is true? a.)

The sample size needs to be increased by 10% so we can apply the Central Limit Theorem. b.)

Nothing can be assumed about the distribution of sample means. c.)

The distribution of sample means has a binomial distribution. d.)

The distribution of sample means is approximately normal.  The central limit theorem tells us that if we randomly draw from the population, then if the sample sizes are sufficiently large, then the sampling distribution will be normal. In general, a sample size of 30 or more is sufficient to guarantee this.

Select the true statement about the relationship between sample size and the standard deviation of distribution of sample means, also known as the standard error. a.)

As sample size decreases, standard error decreases. b.)

Sample size does not have an impact on standard error. c.)

As sample size increases, standard error increases. d.)

As sample size increases, standard error decreases.  Since the sample size (n) is in the denominator in the standard error (mean) calculation, we know that as n increases the standard error decreases.

Which of the following is true about the distribution of sample means? a.)

The mean distribution of sample means is normally greater than the population mean. b.)

The mean distribution of sample means is normally less than the population mean. c.)

The mean distribution of sample means is more variable than population means. d.)

The mean distribution of sample means is the same as the population mean.

 Recall that when you randomly sample to create the sampling distribution of the mean, the mean of the population and the mean of the sampling distribution (of the mean) is the same.

This is the property of unbiasedness. The law of large numbers and central limit theorem illustrate this property.

Jeanette really loves apple-flavored Fruity Tooty candies, but there always seems to be a lot of cherry-flavored candies in each bag. To determine whether this is because cherry candies are so popular or because each bag contains fewer apple candies, Jeanette randomly picks a Fruity Tooty candy from her bag, records its flavor, and places it back in the bag. Each bag contains a mixture of cherry, grape, apple, lemon, and orange flavors.

Which statement about Jeanette's distribution of sample proportions is true? a.)

The distribution of the count of picking a lemon candy can be modeled as approximately normal if Jeanette picks candies 15 times. b.)

The count of drawing an apple candy is not a binomial distribution. c.)

The count of drawing an orange candy has a binomial distribution. d.)

The distribution of the count of picking a cherry candy cannot be modeled as approximately normal if Jeanette picks candies over 100 times.  Recall that proportions can be put into two categories. If we assume that items are independently chosen, this would follow the binomial distribution.

Susan really loves the lemon-flavored Fruity Tooty candies, but there always seems to be a lot of grape-flavored candies in each bag. To determine whether this is because grape candies are so popular or because each bag contains fewer lemon candies, Susan randomly picks a Fruity Tooty candy from her bag, records its flavor, and places it back in the bag. Each bag contains a mixture of cherry, grape, apple, lemon, and orange flavors.

Which statement about Susan's distribution of sample proportions is true? a.)

The distribution of apple candy can be modeled as normal if Susan picks candies over 75 times. b.)

The distribution of lemon candy can be modeled as normal if Susan picks candies 10 times. c.)

The distribution of the count of picking a cherry candy cannot be modeled as approximately normal if Susan picks candies over 200 times. d.)

The sample proportion of drawing an orange candy is not a binomial distribution.  In general the sampling distribution of the proportions follows a binomial distribution. As you let the sample size get larger, we can use a normal approximation to the binomial and note that the sampling distribution converges to a normal distribution.

Theresa really loves orange-flavored Fruity Tooty candies, but there always seems to be a lot of cherry-flavored candies in each bag. To determine whether this is because cherry candies are so popular or because each bag contains fewer orange candies, Theresa randomly picks a

Fruity Tooty candy from her bag, records its flavor, and places it back in the bag. Each bag contains a mixture of cherry, grape, apple, lemon, and orange flavors.

Which statement about Theresa's distribution of sample proportions is true? a.)

The distribution of lemon candy can be modeled as normal if Theresa picks candies 12 times. b.)

The count of drawing an orange candy is not a binomial distribution. c.)

The sample proportion of cherry candies has a binomial distribution. d.)

The distribution of the count of picking an apple candy cannot be modeled as approximately normal if Theresa picks candies over 300 times.  Recall that proportions can be put into two categories. If we assume that items are independently chosen, this would follow the binomial distribution. Theresa can either pick a cherry candy or not pick a cherry candy.

CHALLENEG 2 ackson wants to test an assumption on the number of red candies in a bag of Fruity Tooty's candy. The company claims that five flavors are given an equal distribution in each bag, meaning that the proportion of each color should be 20%. Jackson decides to focus his efforts on examining whether there is a difference in the proportion of red candies than what the company claims.

What should Jackson state for the null hypothesis and alternative hypothesis? a.)

Null hypothesis: Alternative hypothesis: b.)

Null hypothesis: Alternative hypothesis: c.)

Null hypothesis: Alternative hypothesis: d.)

Null hypothesis: Alternative hypothesis:  Since Jackson is testing if the proportion of red is 20%, we can reject that with a larger or smaller percentage. This is a two-tailed test. So the null hypothesis and alternative hypothesis would be: Null hypothesis: p = 0.20 Alternative hypothesis: p ≠ 0.20.

Peter wants to test an assumption on the number of trees that tested positive for a particular infestation. He decides to focus his efforts on examining if more than 12% of trees have been infested.

What should Peter state for the null hypothesis and alternative hypothesis?

a.)

Null hypothesis: Alternative hypothesis: b.)

Null hypothesis: Alternative hypothesis: c.)

Null hypothesis: Alternative hypothesis: d.)

Null hypothesis: Alternative hypothesis:  Since we are focusing on testing if the proportion of is greater than 12%, we can only reject the null hypothesis if the value is larger. This would be a one-tailed right test. So the null hypothesis and alternative hypothesis would be: Null hypothesis: p = 0.12 Alternative hypothesis: p > 0.12.

Megan wants to test her assumption on the amount of sleep students are getting. She decides to focus her efforts on examining if students are getting less than eight hours of sleep.

What should Megan state for the null hypothesis and alternative hypothesis? a.)

Null hypothesis: Alternative hypothesis: b.)

Null hypothesis: Alternative hypothesis: c.)

Null hypothesis: Alternative hypothesis: d.)

Null hypothesis: Alternative hypothesis:  Since we are focusing on testing if the mean is less than 8 hours, we can only reject the null hypothesis if the value is smaller. This would be a one-tailed left test. So the null hypothesis and alternative hypothesis would be: Null hypothesis: μ = 8 Alternative hypothesis: μ < 8.

A school counselor claims that the average number of sleep students get each night is 6 hours.

A researcher has taken a well–designed survey and his sample mean is 8.5 hours and sample standard deviation is 0.25. The sample size is 400.

Which statement is correct? a.)

The sample size should be much more.

b.)

The difference exists due to chance since the test statistic is small. c.)

The result of the survey is not statistically significant. d.)

The result of the survey is statistically significant.  Since we find that 8.5 is much larger than the average 6 hours from a large sample size and relatively small standard deviation, we can conclude there is a statistically significant result.

It is also practically different as well.

A school authority claims that the average height of students is 178 cm. A researcher has taken a well-designed survey and his sample mean is 177.5 cm and the sample standard deviation is 2. The sample size is 25.

Which statement is correct? a.)

The sample mean and population mean is the same. b.)

The difference exists due to chance since the test statistic is small. c.)

The result of the survey is biased. d.)

The result of the survey is statistically significant.  With a very small sample size of 25, a difference of 0.5 cm is most likely due to chance.

A school authority claims that the average percentage marks of students is 68. A researcher has taken a well–designed survey and his sample mean is 64.5 and sample standard deviation is 2. The sample size is 300.

Which statement is correct? a.)

The difference exists due to chance since the test statistic is small. b.)

The result of the survey is statistically significant. c.)

The sample size should be much more. d.)

The result of the survey is not statistically significant.  Since we find that 64.5 is much lower than the average 68 from a large sample size and relatively small standard deviation, we can conclude there is a statistically significant result.

It is also practically different as well.

A spice box manufacturing company is having difficulty filling packets to the required 50 grams. Suppose a business researcher randomly selects 60 boxes, weighs each of them and computes its mean. By chance, the researcher selects packets that have been filled adequately and that is how he gets the mean weight of 50 g, which falls in the "fail to reject" region.

The decision is to fail to reject the null hypothesis even though population mean is NOT actually 50 g.

Which kind of error has the researcher done in this case? a.)

Type II b.)

Type I c.)

Neither d.)

Both  Since the sample evidence provides evidence the null should not be rejected, when in fact the null hypothesis is false, this is called Type II error.

A sugar box manufacturing company is accurately making 100 g packets. Suppose a business researcher randomly selects 100 boxes, weighs each of them and computes its mean. Due to non-random selection or by chance, a researcher selects packets that have not been adequately filled and that is how he gets the mean weight of 98 g, which falls in the rejection region.

The decision is to reject the null hypothesis even though the population mean is actually 100 g.

Which kind of error has the researcher done in this case? a.)

Type II b.)

Type I c.)

Both d.)

Neither  Since the sample evidence provides evidence the null should be rejected, when in fact the null hypothesis is true, this is called Type I error.

A capsule manufacturing company is having difficulty filling capsules to the required 100 milligrams. Suppose a business researcher randomly selects 80 capsules, weighs each of them and computes its mean. By chance, the researcher selects capsules that have been filled adequately and that is how he gets the mean weight of 100 mg, which falls in the "fail to reject" region.

The decision is to fail to reject the null hypothesis even though population mean is NOT actually 100 mg.

Which kind of error has the researcher done in this case? a.)

Both b.)

Type II c.)

Type I d.)

Neither  Since the sample evidence provides evidence the null should not be rejected, when in fact the null hypothesis is false, this is called Type II error.

Which of the following statements is FALSE? a.)

The probability of rejecting the null hypothesis when the null hypothesis is true is called a Type I Error. b.)

Alpha (α) is equal to the probability of making a Type I error. c.)

The power of a hypothesis test is the probability of not making a Type II error. d.)

A smaller sample size would increase the effectiveness of a hypothesis test.  If you decrease the sample size, the ability to detect differences decreases which increases the possibility of error. So the effectiveness of a test would decrease and not increase.

Which of the following statements is FALSE? a.)

The significance level is the probability of making a Type I error. b.)

The probability of rejecting the null hypothesis in error is called a Type I Error. c.)

A larger sample size would increase the power of a significance test. d.) ility to detect differenceExpanding the sample size can decrease the power of a hypothesis test.  If you increase the sample size, the abs increases, which reduces Type II error. So this means the power of the test has increased.

Which of the following statements is FALSE? a.)

Reducing the significance level (α) can increase a test's effectiveness. b.)

A larger sample size would increase the effectiveness of a hypothesis test. c.)

Alpha (α) is equal to the probability of making a Type I error. d.)

Expanding the sample size can increase the power of a hypothesis test.  Recall that alpha is Type I error and corresponds to the threshold we use of acceptable Type I error. So if we increase alpha we are opening ourselves up to more Type I error. This is not an increase in the test's effectiveness.

Susan hypothesizes that the students of a private school will score higher than the general population. Susan records a sample mean equal to 578 and states the hypothesis as vs .

Select the best description for this type of test.

a.)

Two-tailed test b.)

Lower-tailed test c.)

Right-tailed test d.)

Left-tailed test  Since the alternative hypothesis, µ > 572, corresponds to a value that is larger than the null values, we call this a one-tailed upper tail test or right-tailed test.

David hypothesize that the average age of the population of Wyoming is less than 45 years.

David records a sample mean equal to 42 and states the hypothesis as vs .

Select the best description for this type of test. a.)

Two-tailed test b.)

Right-tailed test c.)

Left-tailed test d.)

Upper-tailed test  Since the alternative hypothesis, µ < 45, corresponds to a value that is smaller than the null values, we call this a one-tailed lower tail test or left-tailed test.

The government claims that the average age of Texans is 38 years. Blake hypothesizes that the average age of the population of Texas is not equal to 38 years. Blake records a sample mean equal to 41 and states the hypothesis as vs .

Select the best description for this type of test.

Two-tailed test b.) a.)

Right-tailed test c.)

Left-tailed test d.)

Upper-tailed test  Since the alternative hypothesis, µ ≠ 38, corresponds to values that can be both larger and smaller than the null values, we call this a two-tailed test. When rejecting Ho, we can do so with larger or smaller values that are far enough away from the null value.

Select the correct statement. a.) iven a p-v alue of 0.01, an G d a significance level of 5%, you should reject the null hypothesis. b.)

Given a p-value of 0.05, and a significance level of 3%, you should reject the null hypothesis. c.)

Given a p-value of 0.06, and a significance level of 5%, you should reject the null hypothesis. d.)

Given a p-value of 0.08, and a significance level of 2%, you should reject the null hypothesis.  Recall that our decision rule is to reject Ho if p-value is less than significance level. Since we have 0.01 (p-value) is less than 0.05 (significance level), then we should reject Ho. Select the correct statement. a.)

Given a p-value of 0.06 and a significance level of 4%, you should reject the null hypothesis. b.)

Given a p-value of 0.02 and a significance level of 3%, you should reject the null hypothesis. c.)

Given a p-value of 0.07 and a significance level of 1%, you should reject the null hypothesis. d.)

Given a p-value of 0.04 and a significance level of 3%, you should reject the null hypothesis.  Recall that our decision rule is to reject Ho if p-value is less than significance level. Since we have 0.02 (p-value) is less than 0.03 (significance level), then we should reject Ho. Select the correct statement. a.)

With a p-value of 0.04 and a significance level of 3%, you failed to reject the null hypothesis. b.)

With a p-value of 0.01 and a significance level of 3%, you failed to reject the null hypothesis. c.)

With a p-value of 0.02 and a significance level of 5%, you failed to reject the null hypothesis. d.)

With a p-value of 0.03 and a significance level of 6%, you failed to reject the null hypothesis.  Recall that our decision rule is to reject Ho if p-value is less than the significance level. Since we have 0.04 (p-value) is greater than 0.03 (significance level), then we should FAIL to reject Ho. A recent article claims that the state of Illinois has low tuition rates for its colleges and universities. Cate decides to research the cost of tuition for different colleges in Illinois. The average tuition in Illinois is $28,950 with a standard deviation of $3,470.

Which type of inference test should Cate use? a.)

One-sample t-test b.)

Chi-squared test for goodness-of-fit c.)

One-sample z-test d.)

Two-way ANOVA  Assuming we have a large enough sample, since we are testing one group's mean relative to a hypothesized value, this would be a one-sample z-test.

Tyler wants to determine if an individual's age affects whether or not they like jazz, rock, or country music.

Which type of inference test should Tyler use? a.)

One-sample t-test b.)

Two-way ANOVA c.)

One proportion z-test d.)

Chi-squared test for association/independence  If we break up individuals in age groups and test whether they like or dislike jazz, we would be testing the proportion or counts between the groups. This is testing the association between age and like of jazz, so the appropriate test would be a chi-square test for independence.

A research company is claiming that 75% of college students shop online. Jamila takes a sample of 200 college students and finds that 120 college students shop online.

Which type of inference test should Jamila use? a.)

Two-way ANOVA b.)

One proportion z-test c.)

One-sample t-test d.)

Chi-squared test for homogeneity  Since we are testing a proportion from one group using a sample that is of sufficient size, we would use a one proportion z-test.

CHALLENEG3 The marks scored by students in a geology exam are normally distributed.

If the average score is 122 with a standard deviation of 35, what percentage of students scored below 67? Answers are rounded to the nearest whole percent. a.)

90% b.)

10% c.)

94% d.)

6%  Recall to find the probability we need to convert to a z-score and then go to standard normal chart. We can note:

Looking at a z-table, a z-score of -1.57 corresponds with a percentage of 0.0582 or about 0.06 = 6%.

The cholesterol level of children is normally distributed.

If the average cholesterol level is 194 with a standard deviation of 15, what percentage of children have a cholesterol level lower than 199? Answers are rounded to the nearest whole percent. a.)

74% b.)

26% c.)

37% d.)

63%  Recall to find the probability, we need to convert to a z-score and then go to standard normal chart. We can note: Looking at a z-table, a z-score of 0.33 corresponds with a percentage of 0.6293 or about 0.63 = 63%.

The daily winter temperature in Miami is normally distributed.

Jason sampled 60 smokers who were questioned about the number of hours they sleep each day. Jason wants to test the hypothesis that the smokers need less sleep than the general public which needs an average of 7.5 hours of sleep with a population standard deviation of 0.7 hours.

If the sample mean is 7.2 hours, what is the z-score? Answers are rounded to the hundredths place. a.)

-0.33 b.)

2.67 c.)

-3.32 d.)

0.67  In order to get the z-score we use the formula provided. To make things easier, first calculate the denominator: So 7.2 hours is 3.32 standard deviations below the mean.

If the average winter temperature in Miami is 69°F with a standard deviation of 7°F, what percentage of days would the temperature be above 73°F? Answers are rounded to the nearest whole percent. a.)

28% b.)

43% c.)

57% d.)

72%  Recall to find the probability we need to convert to a z-score and then go to standard normal chart. We can note: Looking at a z-table, a z-score of 0.57 corresponds with a percentage of 0.7157 or about 0.72 = 72%.

However, this is the percentage for the lower distribution. We want the value for the upper, so subtract this from 100%: 100% - 72% = 28%.

Spencer sampled 50 students of a private school who were questioned about their scores in Mathematics. Spencer wants to test the hypothesis that the private school students score better than the general public which has an average of 62 marks with a population standard deviation of 7 marks.

If the sample mean is 65 marks, what is the z-score? Answers are rounded to the hundredths place. a.)

-3.84 b.)

3.03 c.)

-2.56 d.)

1.56  In order to get the z-score we use the formula provided. To make things easier, first calculate the denominator:

Keith sampled 10 private universities in Colorado and recorded the tuition cost. Keith wants to test the hypothesis that the tuition of these 10 universities is more expensive than the national average tuition, which is $29,056 with a population standard deviation of $3,339.

If the sample mean is $31,650, what is the z-score? Answers are rounded to the hundredths place. a.)

2.46 b.)

-0.78 c.)

0.13 d.)

-2.94  In order to get the z-score we use the formula provided. To make things easier, first calculate the denominator: So $31,650 is 2.46 standard deviations above the mean.

A research study claims that 68% of adults drink regularly. Edward conducts a random sample of 200 people and finds that 140 people drink regularly.

Using the formula and data provided, what is the value of the z-test statistic? Answer choices are rounded to the hundredths place. a.)

0.41 b.)

0.61 c.)

0.39 d.)

0.59  In order to get the z-score we use the formula given. To make things easier, first calculate the denominator: This means 140 out of 200 people is 0.61 standard deviations above the population proportion of 68%.

An article claims that 12% of trees are infested by a bark beetle. A random sample of 1,000 trees were tested for traces of the infestation and found that 127 trees were affected.

Using the formula and data provided, what is the value of the z-test statistic? Answer choices are rounded to the hundredths place. a.)

0.25 b.)

0.70 c.)

0.35 d.)

0.50  In order to get the z-score we use the formula given. To make things easier, first calculate the denominator: This means 127 out of 1000 trees is 0.70 standard deviations above the population proportion of 12%.

The Fruity Tooty company claims that the population proportion for each of its five flavors is exactly 20%. Jane counts 92 red candies in a 400-count sample.

Using the formula and data provided, what is the value of the z-test statistic? Answer choices are rounded to the hundredths place. a.)

2.19 b.)

2.24 c.)

1.50 d.)

4.15  In order to get the z-score we use the formula given. To make things easier, first calculate the denominator: This means 92 out of 400 candies is 1.5 standard deviations above the population proportion of 20%.

Select the correct statement. a.)

The critical z-score for a right-tailed test at a 17% significance level is 0.57. b.)

The critical z-score for a two-sided test at a 5% significance level is 1.65. c.)

The critical z-score for a two-sided test at a 3% significance level is 2.17. d.)

The critical z-score for a left-tailed test at a 25% significance level is -0.40.  Recall that when a test is two-sided, we have to split up the percentage for both tails. A 3% significance level means we will have 3% / 2 = 1.5% in each tail. We can start by finding the value in the upper tail as close to (100% - 1.5%) = 98.5% or 0.985 as possible.

In a z-table, the probability of 0.985 corresponds to a z-score of 2.17.

Select the correct statement. a.)

The critical z-score for a two-sided test at a 10% significance level is 1.23. b.)

The critical z-score for a right-tailed test at an 18% significance level is 0.57. c.)

The critical z-score for a left-tailed test at a 14% significance level is -1.08. d.)

The critical z-score for a two-sided test at an 18% significance level is 1.95.  For a left-tailed test, we want to find the value in the lower tail. To do this, we just need to find the closest to 14% or 0.14 in a z-table.

In the z-table, the probability of 0.14 is closest to 0.1401 which corresponds to a z-score of -1.08.

Select the correct statement. a.)

The critical z-score for a two-sided test at a 4% significance level is 1.75. b.)

The critical z-score for a right-tailed test at a 9% significance level is 1.34. c.)

The critical z-score for a two-sided test at a 20% significance level is 0.85. d.)

The critical z-score for a left-tailed test at a 12% significance level is -0.45.  For a right-tailed test, we want to find the value in the upper tail. To do this, we will subtract the significance level from 100%: 100% - 9% = 91% or 0.91.

In a z-table, the probability of 0.91 is closest to 0.9099 which corresponds to a z-score of 1.34.

Spencer randomly samples the out-of-state tuition and fees for 25 public universities and found a mean of $21,032 with a standard deviation of $2,494. Using the alternative hypothesis that µ < $21,706, Spencer found a z test statistic of -1.35.

Given the data provided, what is the p-value of the z-test statistic? Answer choices are rounded to the thousandths place. a.)

0.911 b.)

0.829 c.)

0.089 d.)

0.171  Recall for a p-value, it must be a value that is as extreme and more extreme than the test statistics. If we go to the chart for the z-score and go to the row of -1.3 and to the column

0.05, we can see the value is 0.0885 or 0.089. Since the test is a one-tailed test (note that is less than, "<"), this is the p-value.

Cate randomly samples the out-of-state tuition and fees for 25 public universities and found a mean of $20,340 with a standard deviation of $2,494. Using the alternative hypothesis that µ < $21,706, Cate found a z test statistic of -2.74.

Given the data provided, what is the p-value of the z-test statistic? Answer choices are rounded to the thousandths place. a.)

0.994 b.)

0.006 c.)

0.997 d.)

0.003  Recall for a p-value, it must be a value that is as extreme and more extreme than the test statistics. If we go to the chart for the z-score and go to the row of -2.7 and to the column 0.04, we can see the value is 0.0031 or 0.003. Since the test is a one-tailed test (note that is less than, "<"), this is the p-value.

Kristy randomly samples the out-of-state tuition and fees for 25 public universities and found a mean of $20,876 with a standard deviation of $2,494. Using the alternative hypothesis that µ < $21,706, Kristy found a z test statistic of -1.66.

Given the data provided, what is the p-value of the z-test statistic? Answer choices are rounded to the thousandths place. a.)

0.049 b.)

0.952 c.)

0.903 d.)

0.097  Recall for a p-value, it must be a value that is as extreme and more extreme than the test statistics. If we go to the chart for the z-score and go to the row of -1.6 and to the column 0.06, we can see the value inside the table is 0.0485 or 0.049.

Since the test is a one-tailed test (note that is less than, "<" ), this is the p-value.

What value of z* should be used to construct a 90% confidence interval of a population mean? Answer choices are rounded to the hundredths place. a.)

1.96 b.)

2.58 c.)

1.65

d.)

1.28  Using the z-chart to construct a 90% CI, this means that there is 10% for both tails, or 10% / 2 = 5% for each tail. The lower tail would be at 0.05 and the upper tail would be at (1 - 0.05) or 0.95.

The closest to 0.95 IN THE Z-TABLE is at 0.9505, which corresponds with a z-score of 1.65.

What value of z* should be used to construct a 99% confidence interval of a population mean? Answer choices are rounded to the hundredths place. a.)

2.58 b.)

1.28 c.)

1.96 d.)

1.65  Using the z-chart to construct a 99% CI, this means that there is 1% for both tails, or 1% / 2 = 0.5% or 0.005 for each tail. The lower tail would be at 0.005 and the upper tail would be at (1 - 0.005) or 0.995.

The closest to 0.995 IN THE Z-TABLE is at 0.9951, which corresponds with a z-score of 2.58.

What value of z* should be used to construct a 95% confidence interval of a population mean? Answer choices are rounded to the hundredths place. a.)

3.29 b.)

2.58 c.)

1.96 d.)

1.65  Using the z-chart to construct a 95% CI, this means that there is 5% for both tails, or 5% / 2 = 2.5% for each tail. The lower tail would be at 0.025 and the upper tail would be at (1 - 0.025) or 0.975.

A value of 0.975 found IN THE TABLE corresponds with a z-score of 1.96.

From her purchased bags, Rachel counted 130 red candies out of 520 total candies.

Using a 95% confidence interval for the population proportion, what are the lower and upper limit of the interval? Answer choices are rounded to the thousandths place. a.)

Lower Limit: 0.213 Upper Limit: 0.287 b.)

Lower Limit: 0.247 Upper Limit: 0.287 c.)

Lower Limit: 0.213 Upper Limit: 0.253 d.)

Lower Limit: 0.247 Upper Limit: 0.253  To calculate the limits of a confidence interval, we first need to find the z-score for 95% CI. Recall that 95% CI means there is 5% for both tails, or 2.5% in each tail. This relates to a zscore of 1.96.

Confidence interval can be found using the formula . Plug in the following values: From his purchased bags, Randy counted 120 red candies out of 500 total candies.

Using a 90% confidence interval for the population proportion, what are the lower and upper limits of the interval? Answer choices are rounded to the thousandths place. a.)

Lower Limit: 0.215 Upper Limit: 0.271 b.)

Lower Limit: 0.215 Upper Limit: 0.265 c.)

Lower Limit: 0.209 Upper Limit: 0.265 d.)

Lower Limit: 0.209 Upper Limit 0.271  To calculate the limits of a confidence interval, we first need to find the z-score for 90% CI. Recall that 90% CI means there is 10% for both tails, or 5% in each tail. This relates to a zscore of 1.645.

Confidence interval can be found using the formula . Plug in the following values: From her purchased bags, Rory counted 110 red candies out of 550 total candies.

Using a 90% confidence interval for the population proportion, what are the lower and upper limit of the interval? Answer choices are rounded to the thousandths place. a.)

Lower Limit: 0.166 Upper Limit: 0.241 b.)

Lower Limit: 0.228 Upper Limit: 0.241 c.)

Lower Limit: 0.166 Upper Limit: 0.172

d.)

Lower Limit: 0.172 Upper Limit: 0.228  To calculate the limits of a confidence interval, we first need to find the z-score for 90% CI. Recall that 90% CI means there is 10% for both tails, or 5% in each tail. This relates to a zscore of 1.645.

Confidence interval can be found using the formula . Plug in the following values: A market research company conducted a survey to find the percentage of people who exercise at least 30 minutes every day. Out of 186 persons surveyed, 57 replied that they exercise at least 30 minutes every day.

What is the standard error? Answer choices are rounded to the hundredths place. a.)

0.21 b.)

0.34 c.)

0.03 d.)

1.79  In order to get the standard error of the mean, we can use the following formula: .

Plug in the following values: Correct Sukie interviewed 125 employees at her company and discovered that 21 of them planned to take an extended vacation next year.

What is the standard error? Answer choices are rounded to the thousandths place. a.)

0.033 b.)

0.532 c.)

0.080 d.)

0.015  In order to get the standard error of the mean, we can use the following formula: .

Plug in the following values: A survey was conducted at local colleges around Madison, Wisconsin to find out the average height of a college student. Of 692 students surveyed, 421 replied that they were over 6 feet tall.

What is the standard error? Answer choices are rounded to the hundredths place. a.)

0.24 b.)

12.84 c.)

0.49 d.)

0.02  In order to get the standard error of the mean, we can use the following formula: .

Plug in the following values: Correct I need help with this question CHALLENEG 4 Mary is using a one-sample t-test on the following group: Subject #15: 7.5 hours Subject #27: 6 hours Subject #48: 7 hours Subject #80: 6.5 hours Subject #91: 7.5 hours Subject #82: 8 hours Subject #23: 5.5 hours Select the two TRUE statements. a.)

Mary would use the population standard deviation to calculate a t-distribution. b.)

The t-distribution that Mary uses is taller than a standard distribution. c.)

Mary would use the sample standard deviation to calculate a t-statistic. d.)

The value for the degrees of freedom for Mary's sample population is six. e.)

The t-distribution that Mary uses has skinnier tails than a standard distribution. Answer Rationale Recall with a one-sample t-test, in order to get the t-statistic, we use the sample standard deviation, s. For the t-distribution, we use the degrees of freedom to find the critical value and the degrees of freedom are n-1. So in this case n =7 and df = 7-1 = 6. The t-distribution is a little bit shorter than the standard normal distribution and a little heavier on the tails.

Mark is using a one-sample t-test on the following group: Subject #15: 7.5 hours Subject #27: 6 hours

Subject #48: 7 hours Subject #80: 6.5 hours Subject #91: 6.5 hours Subject #59: 7 hours Select the two TRUE statements. a.)

The value for the degrees of freedom for Mark's sample population is five. b.)

Mark would use the population standard deviation to calculate a t-distribution. c.)

The value for the degrees of freedom for Mark's sample population is six. d.)

The t-distribution that Mark uses has thicker tails than a standard normal distribution. e.)

The t-distribution that Mark uses has thinner tails than a standard distribution. Answer Rationale For the t-distribution, we use the degrees of freedom to find the critical value and the degrees of freedom are n-1. So in this case n = 6 and df = 6-1 = 5. Also, recall that the t-distribution for lower degrees of freedom has shorter, fatter tails than the standard normal distribution.

Miriam is using a one-sample t-test on the following group: Subject #15: 6.5 hours Subject #27: 5 hours Subject #48: 6 hours Subject #80: 7.5 hours Subject #91: 5.5 hours Select the two TRUE statements. a.)

The value for the degrees of freedom for Miriam's sample population is five. b.)

The t-distribution that Miriam uses is shorter and has thicker tails than a normal distribution. c.)

The value for the degrees of freedom for Miriam's sample population is four. d.)

The t-distribution that Miriam uses is taller and has thinner tails than a normal distribution. e.)

Miriam needs to use a t-test when the standard deviation is known. Answer Rationale For the t-distribution, we use the degrees of freedom to find the critical value and the degrees of freedom are n-1. So in this case n =5 and df = 5-1 = 4. Also, recall that the t-distribution for lower degrees of freedom has shorter, fatter tails than the standard normal distribution.

Jessica knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help her see if college students are different from the adult population.

Jessica tabulated that her sample of 101 students got an average of 6.9 hours of sleep each night, with a standard deviation of 2.5.

Using the formula and data provided, what is the t-statistic that Jessica calculates? Answer choices are rounded to the hundredths place. a.)

-4.42 b.)

6.58 c.)

0.28 d.)

-1.10  To find the value of the t-statistic, it can be easiest to evaluate the denominator: Julie knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help her see if college students are different from the adult population.

Julie tabulated that her sample of 101 students got an average of 7.1 hours of sleep each night, with a standard deviation of 2.48.

Using the formula and data provided, what is the t-statistic that Julie calculates? Answer choices are rounded to the hundredths place. a.)

6.78 b.)

-3.64 c.)

2.61 d.)

0.24  To find the value of the t-statistic, it can be easiest to evaluate the denominator: Correct Go to the Next Concept I need help with this question Jacob knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help him see if college students are different from the adult population.

From a sample of 101 students, Jacob tabulated that his sample of students got an average of 7.3 hours of sleep each night, with a standard deviation of 2.6.

Using the formula and data provided, what is the t-statistic that Jacob calculates? Answer choices are rounded to the hundredths place. a.)

3.70 b.)

7.72 c.)

-2.71 d.)

6.99  To find the value of the t-statistic, it can be easiest to evaluate the denominator: Marcus collected 30 responses to a survey and used a two-tailed test to establish a significance level of 0.05 and calculate a 95% confidence interval.

Using the t-table (located in the tutorial), what is the critical t-value used for calculating a 95% confidence interval with 29 degrees of freedom? Answer choices are rounded to the thousandths place. a.)

2.150 b.)

2.045 c.)

1.699 d.)

2.462  If the sample is n = 30, then the df = n-1 = 29. Recall that a 95% confidence interval would have 5% for the tails, so 2.5% for each tail. So if we go to the row where df = 29 and then 0.025 for the tail probability, this gives us a value of 2.045. We can also use the last row and find the corresponding confidence level.

Thomas collected 29 responses to a survey and used a two-tailed test to establish a significance level of 0.20 and calculate a 90% confidence interval.

Using the t-table (located in the tutorial), what is the critical t-value used for calculating a 90% confidence interval with 28 degrees of freedom? Answer choices are rounded to the thousandths place. a.)

1.701 b.)

1.311 c.)

1.699 d.)

2.467  If the sample is n = 29, then the df = n-1 = 28. Recall that a 90% confidence interval would have 10% for the tails, so 5% for each tail. So if we go to the row where df = 28 and then 0.05 for the tail probability, this gives us a value of 1.701. We can also use the last r Ashley collected 25 responses to a survey and used a two-tailed test to establish a significance level of 0.02 and calculate a 98% confidence interval.

Using the t-table (located in the tutorial), what is the critical t-value used for calculating a 98% confidence interval with 24 degrees of freedom? Answer choices are rounded to the thousandths place. a.)

1.699 b.)

2.492 c.)

2.045 d.)

1.311  If the sample is n = 25, the the df = n-1 = 24. Recall that a 98% confidence interval would have 2% for the tails, so 1% for each tail. So if we go to the row where df = 24 and then 0.01 for the tail probability, this gives us a value of 2.492. We can also use the last row and find the corresponding confidence level.

The average age of Iowa residents is 37 years. Amy believes that the average age in her hometown in Iowa is not equal to this average and decided to sample 30 citizens in her neighborhood.

Using the alternative hypothesis that µ ≠ 531, Amy found a t-test statistic of 1.311.

What is the p-value of the test statistic? Answer choices are rounded to the hundredths place. a.)

0.10 b.)

0.20 c.)

0.07 d.)

0.05  In order to find the p-value we first need the degrees of freedom because we are running a ttest. The df = n-1 = 30-1 = 29. If Amy is using the alternative hypothesis, µ ≠ 531, this is a two-tailed test. In order to find the p-value, we simply need to find the probability of being as extreme or more extreme as our current test statistic in one-tail and then multiply it by 2.

Another way to describe this is it is the tail probability/area. If we go to the df row of 29 and then scan to the right we see 1.311 is in the 0.10 column. So if we multiply this by 2, we find 0.20 or 20% in both of the tails. So the p-value is 0.20.

The average weight of a cereal box is 25 ounces. William believes the cereal boxes at his local grocery store are less than the claimed average and decided to sample 15 cereal boxes from this grocery store.

Using the alternative hypothesis that µ < 25, William found a t-test statistic of -2.624.

What is the p-value of the test statistic? Answer choices are rounded to the hundredths place. a.)

0.05 b.)

0.07

c.)

0.01 d.)

0.02  In order to find the p-value, we first need the degrees of freedom because we are running a ttest. The df = n-1 = 15-1 = 14. Since this is a one-tailed test, in order to find the p-value we simply need to find the probability of being as extreme or more extreme as our current test statistic. Another way to describe this is it is the tail probability/area. If we go to the df row of 14 and then scan to the right we see 2.624 is in the 0.01 column. This indicates there is 0.01 or 1% in the tail. So the p-value is 0.01.

The average SAT score in California is 531. A private high school believes their students scored significantly better on the SAT than the state average and decided to sample 20 of their students.

Using the alternative hypothesis that µ > 531, the high school found a t-test statistic of 1.729.

What is the p-value of the test statistic? Answer choices are rounded to the hundredths place. a.)

0.95 b.)

0.05 c.)

0.19 d.)

0.20  In order to find the p-value we first need the degrees of freedom because we are running a ttest. The df = n-1 = 20-1 = 19. Since this is a one-tailed test, in order to find the p-value, we simply need to find the probability of being as extreme or more extreme as our current test statistic. Another way to describe this is it is the tail probability/area. If we go to the df row of 19 and then scan to the right we see 1.729 is in the 0.05 column. This indicates there is 0.05 or 5% in the tail. So the p-value is 0.05.

Correct Aaron sampled 101 students and calculated an average of 6.5 hours of sleep each night with a standard deviation of 2.14. Using a 96% confidence level, he also found that t* = 2.081.

A 96% confidence interval calculates that the average number of hours of sleep for working college students is between __________. Answer choices are rounded to the hundredths place. a.)

6.08 to 6.94 b.)

6.08 to 6.92 c.)

6.06 to 6.94 d.)

6.46 to 6.92  Using the given formula and the information provided: Morgan sampled 101 students and calculated an average of 6.5 hours of sleep each night with a standard deviation of 2.14. Using a 90% confidence level, she also found that t* = 1.660.

A 90% confidence interval calculates that the average number of hours of sleep for working college students is between __________ hours. Answer choices are rounded to the hundredths place. a.)

6.46 and 6.85 b.)

6.08 and 6.92 c.)

6.46 and 6.54 d.)

6.15 and 6.85  Using the given formula and the information provided: Correct Go to the Next Concept ackson sampled 101 students and calculated an average of 6.5 hours of sleep each night with a standard deviation of 2.14. Using a 95% confidence level, he also found that t* = 1.984.

A 95% confidence interval calculates that the average number of hours of sleep for working college students is between __________ hours. Answer choices are rounded to the hundredths place. a.)

6.46 and 6.54 b.)

6.15 and 6.94 c.)

6.08 and 6.92 d.)

6.46 and 6.92  Using the given formula and the information provided: Correct The daily high temperatures, in degrees Fahrenheit, of Des Moines for one week were: 64.5 64

66.5 64 62.5 61 63 Using the data above, what is the standard error of the sample mean? Answer choices are rounded to the hundredths place. a.)

9.09 b.)

0.25 c.)

1.53 d.)

0.65  To find the standard error, we first need to find the sample standard deviation. To quickly find this, use the function =STDEV.S in Excel.

The standard deviation of the s = 1.73 with n = 7. So the standard error is equal to: Correct Jacob was researching the number of hours the average adult sleeps each night. He collected data from five individuals: Subject #1: 6 hours Subject #2: 6.5 hours Subject #3: 6 hours Subject #4: 7.5 hours Subject #5: 5 hours Using the data above, what is the standard error of the sample mean? Answer choices are rounded to the hundredths place. a.)

6.10 b.)

0.41 c.)

0.74 d.)

0.15  To find the standard error, we first need to find the sample standard deviation. To quickly find this, use the function =STDEV.S in Excel.

The standard deviation of the s = 0.91 with n = 5. So the standard error is equal to:

The weights, in pounds, of eight students in a class are: 128 193 166 147 202 183 181 158 Using the data above, what is the standard error of the sample mean? Answer choices are rounded to the hundredths place. a.)

3.10 b.)

0.11 c.)

8.76 d.)

6.85  To find the standard error, we first need to find the sample standard deviation. To quickly find this, use the function =STDEV.S in Excel.

The standard deviation of the sample is = 24.77 with n = 8. So the standard error is equal to: Select the true statement regarding the F-statistic of an ANOVA test. a.)

The smaller the F value, the greater the variation between individual samples. b.)

The larger the F value, the more likely it is to reject the null hypothesis. c.)

The larger the F value, the more likely it is to fail to reject the null hypothesis. d.)

The smaller the F value, the more likely it is to reject the null hypothesis.  Recall that an F-distribution is a one-tailed distribution to the right, so larger values mean more extreme. Larger values would imply a higher likelihood of rejecting the null hypothesis depending on the level of alpha your test is.

Incorrect Go to the Next Question CHALLENEG 5 What is NOT assumed before starting an ANOVA test? a.)

The observations are independent and from a random sample. b.)

The population standard deviations are similar. c.)

Populations are normally distributed. d.)

The population means are the same.  A one way ANOVA is used to test the differences in population mean values. So we don't assume they are the same. We are testing to see if there is evidence if they are the same or different.

Correct Go to the Next Concept How is the F-statistic in an ANOVA test calculated? a.)

The variance between the samples plus the variance within the samples b.)

The variance between the samples divided by the variance within the samples c.)

The variance between the samples multiplied by the variance within the samples d.)

The variance between the samples minus the variance within the samples  In order to get the F stat, we take the mean squares between the samples divided by the mean squared variation within the samples. If the between samples variation is much larger, it is evidence that the groups are different.

Which of the following is an example of a two-way ANOVA test? a.)

Analyzing the number of hours that freshman, sophomores, juniors, and seniors study in a semester as well as the number of hours of sleep each group gets at night. b.)

Comparing the number of hours of sleep for freshman, sophomores, juniors, and seniors. c.)

Comparing the number of hours that freshman and seniors study during a semester. d.)

Studying the number of hours of sleep for freshmen, sophomores, and juniors.  Recall that a two-way ANOVA examines differences in mean of two or more factors between three or more populations. Since we are exploring hours by grade level (factor 1) and by sleep hours (factor 2) among 4 different populations (freshman, sophomore, juniors and seniors), this is an example of a two-way ANOVA. Which of the following is an example of a one-way ANOVA test? a.)

Studying the number of hours slept as well as the average GPAs for middle school, high school, and college students. b.)

Comparing the number of hours of sleep between children and teenagers. c.)

Studying the number of hours of sleep for teenagers, adults, and senior citizens. d.)

Comparing the monthly incomes of residents who live in urban and suburban areas.

 Recall that a one-way ANOVA examines differences in mean of a single factor between three or more populations. Since we are only exploring hours of sleep (single factor) between 3 different populations (teenagers, adults, and senior citizens), this is an example of a one-way ANOVA. Which of the following is an example of a two-way ANOVA test? a.)

Comparing the number of hours of sleep for children, teenagers, adults, and senior citizens. b.)

Studying the monthly income of residents who live in rural, urban, and suburban areas. c.)

Studying the number of hours of sleep as well as primary residence (rural, urban, or suburban) for children, teenagers, adults, and senior citizens. d.)

Comparing the number of hours of sleep between children and adults.  Recall that a two-way ANOVA examines differences in mean of two or more factors between three or more populations. Since we are exploring hours of sleep (factor 1) and residence (factor 2) between 4 different populations (children, teenagers, adults and senior citizens), this is an example of a two-way ANOVA. Kathy recorded the results for a sample of 502 candies.

The candy company claims that the distribution of each color is exactly 20%.

Select the observed and expected frequencies for the red candies. a.)

Observed: 100.4 Expected: 120 b.)

Observed: 120 Expected: 75.3 c.)

Observed: 106 Expected: 75.3 d.)

Observed: 120 Expected: 100.4  To get the observed frequency, we simply look at the value in the column. In this case, it is 120 for the red group.

To get the expected frequency, we use the formula: Correct Go to the Next Concept Kaitlin recorded the results for a sample of 536 candies.

The candy company claims that the distribution of each color is exactly 20%.

Select the observed and expected frequencies for the yellow candies. a.)

Observed: 123 Expected: 80.4 b.)

Observed: 145 Expected: 107.2 c.)

Observed: 107.2 Expected: 14 d.)

Observed: 145 Expected: 80.4  To get the observed frequency, we simply look at the value in the column. In this case, it is 145 for the yellow group.

To get the expected frequency, we use the formula: Karen recorded the results for a sample of 518 candies.

The candy company claims that the distribution of each color is exactly 20%.

Select the observed and expected frequencies for the green candies. a.)

Observed: 103.6 Expected: 93 b.)

Observed: 93 Expected: 103.6 c.)

Observed: 93 Expected: 93.2 d.)

Observed: 72 Expected: 93.2  To get the observed frequency, we simply look at the value in the column. In this case, it is 93 for the green group.

To get the expected frequency, we use the formula: Robert recorded the number of calls he made at work during the week: Monday: 20 Tuesday: 12 Wednesday: 10

Thursday: 18 He expected to make 15 calls each day. To determine whether the number of calls follows a uniform distribution, a chi square test for goodness of fit should be performed (alpha = 0.05).

Using the data above, what is the chi-square test statistic? Answer choices are rounded to the hundredths place. a.)

0.42 b.)

4.54 c.)

3.75 d.)

0.67  Using the above formula we can note the chi-square test statistic is: Cindy recorded the number of customers who visited her new shop during the week: Wednesday: 18 Thursday: 20 Friday: 25 Saturday: 17 She expected to have 20 customers each day. To determine whether the number of customers follows a uniform distribution, a chi square test for goodness of fit should be performed (alpha = 0.10).

Using the data above, what is the chi-square test statistic? Answer choices are rounded to the hundredths place. a.)

1.9 b.)

0.37 c.)

3.51 d.)

0.55  Using the above formula we can note the chi-square test statistic is: Betty recorded the number of visitors at her art gallery during the week: Tuesday: 28 Wednesday: 20

Thursday: 18 Friday: 14 She expected to have 20 visitors each day. To determine whether the number of visitors follows a uniform distribution, a chi square test for goodness of fit should be performed (alpha = 0.10).

Using the data above, what is the chi-square test statistic? Answer choices are rounded to the hundredths place. a.)

0.83 b.)

5.2 c.)

0.20 d.)

12.5  Using the above formula we can note the chi-square test statistic is: A table represents a possibility of association between eye color and hair color.

In order to determine if there is a significant difference between eye color and hair color, the chi square test for association and independence should be performed.

What is the expected frequency of Blue Eyes and Black Hair? a.)

19 b.)

18 c.)

17 d.)

20  In order to get the expected count we use the formula: The row total of blue eyes is 18 + 19 = 37.

The column total of black hair is 15 + 19 = 34.

The total of the whole table is 22 + 15 + 18 +19 = 74.

So the expected frequency of blue eyes and brown hair is: A table represents a possibility of association between blood type and eye color:

In order to determine if there is a significant difference between blood type and eye color, a chi square test for homogeneity should be performed.

What is the expected frequency of Group B and Green Eyes? Answer choices are rounded to the nearest hundredth. a.)

15.67 b.)

16.11 c.)

18.50 d.)

15.83  In order to get the expected count we use the formula: The row total of Group B is 13 + 12 + 22 = 47.

The column total of Green Eyes is 19 + 12 + 16 + 17 = 64.

The total of the whole table can be found by adding all values, or 190.

So the expected frequency of Green eyes and Group B is: A table represents a possibility of association between hair color and eye color.

In order to determine if the eye color of blue and blonde hair colors differ significantly, a chi square test for homogeneity should be performed.

What is the expected frequency of Blonde Hair and Blue Eyes? Answer choices are rounded to the nearest hundredth. a.)

25.50 b.)

27.67 c.)

28.41 d.)

27.25  In order to get the expected count we use the formula: The row total of blonde hair is 25 + 27 + 31 = 83.

the column total of blue eyes is 25 + 26 = 51.

The total of the whole table can be found by adding all values, or 149.

The expected frequency of blonde hair and blue eyes is: A table represents a possibility of association between eye color and hair color.

What is the degrees of freedom? a.)

2 b.)

1 c.)

3 d.)

4  In order to get degrees of freedom, we use the following formula: A table represents a possibility of association between hair color and eye color.

What is the degrees of freedom? a.)

2 b.)

1 c.)

6 d.)

4  In order to get degrees of freedom, we use the following formula: Correct A table represents a possibility of association between blood type and eye color:

What is the degrees of freedom? a.)

6 b.)

2 c.)

12 d.)

7  In order to get degrees of freedom, we use the following formula: Correct