In Exercises 5–20, assume that the two samples are independent...

Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)

Bicycle Commuting A researcher used two different bicycles to commute to work. One bicycle was steel and weighed 30.0 lb; the other was carbon and weighed 20.9 lb. The commuting times (minutes) were recorded with the results shown below (based on data from “Bicycle Weights and Commuting Time,” by Jeremy Groves, British Medical Journal).

b. Construct the confidence interval suitable for testing the claim in part (a).

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says a sports scientist is investigating whether there is a difference in average sprint times and seconds between athletes who use a new training program to those who use a standard program. The following data were collected, and we're given a table that has 4 columns. In the first column, we have the training program. For the second column, we have the sample size in. For the third column, we have the mean sprint time X bar given in seconds. And for the last column, we have the standard deviation S, and that's also given in seconds. So for the first row, we have under the training program, the new training program. It has a sample size of 18, a mean sprint time of 11.2, and a standard deviation of 0.6. And for the next row, we have our standard program, and it has a sample size of 21, a mean sprint time of 11.5, and a standard deviation of 0.7. Assume that the two samples are independent simple random samples selected from a normally distributed population and do not assume that the population standard deviations are equal. Use degrees of freedom equal to the smaller of N1 minus 1 and N2 minus 1. Construct a 95% confidence interval for the difference in mean sprint times, where the difference is defined as the mean for the new program minus the mean for the standard program. So, in this problem, we're asked to construct a 95% confidence interval for the difference in mean sprint times. So recall your formula for the confidence interval for the difference in means, assuming that we have independent samples and unequal variances. So for our formula, we'll have the quantity of X 1 minus X2. Plus or minus T star. Multiplied by the square root. Of the quantity of. S 12 divided by N1 plus S2 2 divided by N2 in quantity where group 1 is going to be the new training program and group 2 is going to be the standard training program, and in this formula T star is going to be our critical value. So we were given the mean sprint times as well as standard deviations for both groups. So the only thing we really need to calculate in order to use this formula is going to be our critical value. And for that, we'll first need to determine the number of degrees of freedom that we have. So we'll label that as DF, and this is going to be equal to, and we're told in the problem that this going to be the smaller of N1 minus 1 and N2 minus 1. So this is going to be equal to the minimum of N1 minus 1 and N2 minus 1. So this is going to be equal to the minimum. Of for N1, we have a sample size of 18, so we'll have 18 minus 1, and then for N2, we have a sample size of 21, so we'll have 21 minus 1. And so that means I'm really looking for the minimum. Of 17 and 20, and the smaller of those two values is going to be equal to 17. Now the other thing that we need to find is our value for alpha. And so since we're looking for a 95% confidence interval, that means that alpha is going to be equal to 1 minus 0.95. So here we just converted the 95% into a decimal. And so that means that alpha is going to be equal to 0.05. However, when we look up at the table, we're actually going to need the quantity of alpha divided by 2. So that's going to be equal to 0.05 divided by 2, which is going to be equal to 0.025. And so now if we look up our critical value in our tables using 1740 degrees of freedom and 0.025 for alpha divided by 2, we see that T is going to be equal to 2.110. So now we have all the information that we need to calculate our confidence interval. So, substituting in the values into our formula will have the quantity of X1, which is 11.2, minus X2, which is 11.5, in quantity, you'll have plus or minus. 2.110 multiplied by the square root of the quantity. For S1, that is going to be 0.6, and that gets squared, and that's divided by N1, which is 18. Then we'll have plus. For S2, we have 0.7, and that quantity is going to be squared. And that gets divided by into, which is 21 in quantity. And so, evaluating this expression. For the first half, we'll have 11.2 minus 11.5, which is going to be minus 0.3. Then we'll have plus or minus, and when we evaluate the second half of this expression, this is going to be approximately equal to. 0.4. And so here we just rounded to the same number of decimal places that our difference of our two means is rounded to. So if we want to write this in the interval notation, that means that our lower limit is going to be -0.3 minus 0.4, which is going to be -0.7, and this actually has units of seconds since our time was measured in seconds. And then for our upper bound, we'll have -0.3 plus 0.4, which gives me 0.1 seconds. And so, our 95% confidence interval is going to be the interval from -0.7 seconds to 0.1 seconds. That is going to be the answer that we're looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Confidence Interval

Independent Samples

T-distribution

Watch next