Statistics Notes Confidence Intervals for the Mean – Small Samples

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Statistics Confidence Intervals for the Mean – Small Samples


According to the central limit theorem, the sampling distribution of a statistic (like a sample mean) will follow a normal distribution, as long as the sample size is sufficiently large. But sample sizes are sometimes small, and often we do not know the standard deviation of the population. When either of these problems occurs, statisticians rely on the distribution of the t statistic (also known as the t-score), whose values are given by: xď€­ď ­ t s/ n


where x is the sample mean, Âľ is the population mean, s is the standard deviation of the sample, and n is the sample size. The distribution of the t statistic is called the t distribution or the Student t distribution. The t-distribution or Student's distribution arises in the problem of estimating the mean of a normally distributed population when the sample size is small, as well as when (as in nearly all practical statistical work) the population standard deviation is unknown and has to be estimated from the data.


1. The t-distribution is bell shaped and symmetric about the mean. The tails in the t-distribution are thicker than those in the standard normal distribution. These tails are reduced as the size of the sample increase. When the sample size is greater than 30, the t-distribution curve approaches the normal distribution. 2. The t-distribution is a family of curves, each determine by a parameter called the degrees of freedom. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size. Degrees of Freedom = d.f. = n – 1



3. As the degrees of freedom increase, the tdistribution approaches the normal distribution. After 30 degrees of freedom the t-distribution is very close to the standard normal z-distribution. Recall that a confidence interval for a normal distribution curve 95% of all z values lie between 1.9600 and -1.9600. Roughly two deviations above and below 0.


In a t-distribution with only 5 degrees of freedom, meaning a sample size of 6, 95% of all t values will lie between 2.571 and -2.571.


But with 20 degrees of freedom, meaning a sample size of 21, 95% of all t values will lie between 2.086 and -2.086.


Using the graphing utility.

In a t-distribution with 20 degrees of freedom, 95% of all t values will lie between 2.0860 and -2.0860. With only 5 degrees of freedom, 95% of all t values will lie between 2.5706 and -2.5706.

The value is negative. Just remember the critical value for t would be positive.


Guidelines: Constructing a confidence interval for the mean using t-distribution. 1. Find the sample mean and its standard deviation. 2. Identify the degrees of freedom, the level of confidence desired, and the critical value tc . s 3. Find the maximum error of estimate E. E  tc n

4. Find the left and right endpoints and form the confidence interval. x ď‚ą E

5. Finally make sure you word your response correctly.


Example. Confidence interval In many product quality tests you will deliberately destroy the product to find some information about it. If the product is expensive (such as cars or major appliances) you don't want to use a large sample size. Suppose you select a sample of 15 cars and subject them to crash tests and analyze your results for the dollar amount of damage done. If the mean dollar amount of damage is $15,789 with a standard deviation of $4782.50, find a 99% confidence interval for the mean.


The sample size is 15. The degrees of freedom are 14 The sample mean is 15789 The sample standard deviation is 4782.50 99% confidence interval. s 4782.50  2.9768   3675.8604 Find E. E  tc n 15



The interval will be the sample mean plus and minus this error. x  E  15789  3675.8604  12113.1396 x  E  15789  3675.8604  19464.8604

 $12113.14,$19464.86

With 99% confidence we can say the interval ($12113.14, $19464.86) will contain the mean dollar amount of damage.



Example. Confidence interval. A chemical process has produced the following daily yields for the past week: 785, 805, 790, 793, and 802 tons. a. Construct a 95% confidence interval for the population mean Âľ. b. What assumptions did you make? The sample size is 5. The degrees of freedom are 4. The assumption we will make is that the population is normal or approximately normal.


With 95% confidence we can say the interval (784.65, 805.35) tons will contain the mean amount of daily yield.


The following flowchart details when to use the normal distribution to construct a confidence interval and when to use a t-distribution.


Example. Decide. You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost? Explain your reasoning.


SOLUTION Because the population is normally distributed and the population standard deviation is known, you should use the normal distribution.

I am 95% confident that the interval ($170024, $191976) contains the population mean construction cost of newly constructed houses.


Example. Decide. You randomly select 18 adult male athletes and measure the resting heart rate of each. The sample mean heart rate is 64 beats per minute with a sample standard deviation of 2.5 beats per minute. Assuming the heart rates are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 90% confidence interval for the mean heart rate? Explain your reasoning.


SOLUTION Because the population is normally distributed and the sample standard deviation is known instead of the population standard deviation, you should use the t-distribution.

I am 90% confident that the interval (62.975, 65.025) contains the population heart rate of adult male athletes.



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