Question

In: Statistics and Probability

The upcoming championship high school football game is a big deal in your little town. The...

The upcoming championship high school football game is a big deal in your little town. The problem is, it is being played in the next biggest town, which is two hours away! To get as many people as you can to attend the game, you decide to come up with a ride-sharing app, but you want to be sure it will be used before you put all the time in to creating it. You determine that if more than three students share a ride, on average, you will create the app.

You conduct simple random sampling of 20 students in a school with a population of 300 students to determine how many students are in each ride-share (carpool) on the way to school every day to get a good idea of who would use the app. The following data are collected:

  • 6 5 5 5 3 2 3 6 2 2
  • 5 4 3 3 4 2 5 3 4 5

You and your partner will construct a 90% confidence interval and 95% confidence interval for the mean number of students who share a ride to school, then you will compare and interpret the results.

Part A: Decide which partner will construct each confidence interval, and then state the parameter and check the conditions. (Individual work)

Part B: Construct the confidence interval. Be sure to show all your work, including the degrees of freedom, critical value, sample statistics, and an explanation of your process. (Individual work)

Part C: Interpret, share, and compare. Interpret the meaning of the confidence interval, compare your results with your partner, and describe the difference in the width of the interval and the margin of error. (Collaborative work)

Part D: Use your findings to explain whether you should develop the ride-share app for the football game.

Solutions

Expert Solution

I will construct the 90% confidence interval. Please post the question again for the 95% confidence interval.

The data is:

Mean of the data =

The standard deviation of the data =

The t-critical value for df = 20 - 1 = 19 and is 1.73.

The 90% confidence interval is:

= 3.850 1.73*

= 3.329, 4.371

The 90% confidence interval for the mean number of students who share a ride to school is between 3.329 and 4.371.

We are 90% confident that the mean number of students who share a ride to school is between 3.329 and 4.371.


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