Confidence Intervals for Means and Proportions for AP Statistics (page 2)

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By — McGraw-Hill Professional
Updated on Feb 4, 2011

solution: The problem states that the sample is an SRS. The large sample size justifies the construction of a one-sample confidence interval for the population mean. For a 95% confidence interval with df = 81 – 1 = 80, we have, from Table B, t* =1.990. We have 12.5 ± (1.99) = (11.64, 13.36)

If the problem had stated that n = 80 instead of 81, we would have had df = 80 – 1 = 79. There is no entry in Table B for 79 degrees of freedom. In this case we would have had to round down and use df = 60, resulting in t* = 2.000 and an interval of 12.5 ± (2.00) = (11.63, 13.37) the difference isn't large, but the interval is slightly wider. (For the record, we note that the value of t* for df = 79 is given by the TI-84 as invT(0.975,79) = 1.99045.)

You can use the STAT TESTS TInterval function on the TI-83/84 calculator to find a confidence interval for a population mean (a confidence interval for a population mean is often called a "one-sample" t interval). It's recommended that you show how the interval was constructed as well as reporting the calculator answer. And don't forget to show that the conditions needed to construct the interval are present.

example: Interpret the confidence interval from the previous example in the context of the problem.

solution: We are 95% confident that the true mean number of unoccupied seats is between 11.6 and 13.4 seats. (Remember that we are not making any probability statement about the particular interval we have constructed. Either the true mean is in the interval or it isn't.)

For large sample confidence intervals utilizing z-procedures, it is probably worth memorizing the critical values of z for the most common C levels of 0.90, 0.95, and 0.99. They are:

example: Brittany thinks she has a bad penny because, after 150 flips, she counted 88 heads. Find a 99% confidence interval for the true proportion of heads. Do you think the coin is biased?

solution: First we need to check to see if using a z interval is justified.

Because n and n(1 – ) are both greater than or equal to 5, we can construct a 99% z interval for the population proportion:

We are 99% confident that the true proportion of heads for this coin is between 0.484 and 0.69. If the coin were fair, we would expect, on average, 50% heads. Since 0.50 is in the interval, it is a likely population value for this coin. We do not have strong evidence that Brittany's coin is bad.

Generally, you should use t procedures for one- or two-sample problems (those that involve means) unless you are given the population standard deviation(s) and z-procedures for one- or two-proportion problems.

example: The following data were collected as part of a study. Construct a 90% confidence interval for the true difference between the means (μ1 – μ2). Does it seem likely the differences in the sample indicate that there is a real difference between the population means? The samples were SRSs from independent, approximately normal, populations.

solution: The relatively small values of n tell us that we need to use a two-sample t interval. The conditions necessary for using this interval are given in the problem: SRSs from independent, approximately normal, populations. Using the "conservative" method of choosing the degrees of freedom:

We are 90% confident that the true difference between the means lies in the interval from 0.227 to 5.25. If the true difference between the means is zero, we would expect to find 0 in the interval. Because it isn't, this intervalprovides evidence that there might be a real difference between the means.

If you do the same exercise on your calculator (STAT TESTS 2-SampTInt), you get (0.302, 5.178) with df = 35.999. This interval is narrower, highlighting the conservative nature of using df = min{ n1 – 1, n2 – 1 }. Also, note the calculator calculates the number of degrees of freedom using

example: Construct a 95% confidence interval for p1p2 given that n1 = 180, n2 = 250, 1 = 0.31, 2 = 0.25. Assume that these are data from SRSs from independent populations.

solution: 180(0.31) = 55.8, 180(1 – 0.31) = 124.2, 250(0.25) = 62.5, and 250(0.75) = 187.5 are all greater than or equal to 5 so, with what is given in the problem, we have the conditions needed to construct a two-proportion z interval.

Practice problems for these concepts can be found at:

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