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A bank wonders if omitting the annual credit card fee for customers who charge at least \(2400in a year will increase the amount charged on its credit cards. The bank makes this offer to an SRS of 200of its credit card customers. It then compares how much these customers charge this year with the amount that they charged last year. The mean increase in the sample is \)332, and the standard deviation is$108.

a. Construct and interpret a 99%confidence interval for the true mean increase.

b. Based on the interval from (a), can you conclude that dropping the annual fee would cause an increase in the average amount spent by this bank’s credit card customers? Why or why not?

Short Answer

Expert verified

Part (a) We are 95%confident that the true mean increases is between 311.8956and352.1044.

Part (b) No, we cannot conclude.

Step by step solution

01

Part (a) Step 1. Given information

It is given:

x¯=332n=200s=108c=0.99

02

Part (b) Step 2. Calculation

The degree of freedom will be:

df=n-1=200-1=199

Now, the value of t will be:

tα/2=2.626

Now, the confidence interval will be:

x¯-tα/2×sn=332-2.626×108200=311.8956=x¯+tα/2×sn=332+2.626×108200=352.1044

Thus we conclude that we are 99% confident that the true mean increases is between311.8956and352.1044.

03

Part (b) Step 1. Explanation

A completely randomized experiment randomly assigns all subjects to a group.

In this case, the experiment is not completely randomized since we use the subjects before and after the treatment. Thus the experiment is not a completely randomized experiment.

We cannot prove causation if we do not use a completely randomized experiment because it is possible that the causation of the difference is another variable.

For example, inflation could be the cause because everything you buy will be slightly more expensive the next year due to inflation and thus the increase in the amount charged could be due to inflation.

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Most popular questions from this chapter

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select10Variety A and10Variety B tomato plants. Then the researchers divide in half each of10small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The10differences (Variety A − Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is x-=0.343051526=0.200=20%x-A-B=0.34and the standard deviation of the differences is s A-B=0.833051526=0.200=20%=sA-B=0.83.LetμA-B=3051526=0.200=20%μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

The P-value for a test of H0: μA−B=03051526=0.200=20%versus Ha: μA−B≠0 is 0.227. Which of the following is the

correct interpretation of this P-value?

a. The probability that μA−B is0.227.

b. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34is0.227.

c. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34or greater is0.227.

d. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference greater than or equal to0.34or less than or equal to −0.34is0.227.

e. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is not 0, the probability of getting a sample mean difference greater than or equal to 0.34or less than or equal to −0.34is0.227.

There are two common methods for measuring the concentration of a pollutant in fish tissue. Do the two methods differ, on average? You apply both methods to each fish in a random sample of 18carp and use

a. the paired t test for μdiff3051526=0.200=20.0%μdiff.

b. the one-sample z test for p.

c. the two-sample t test for μ1-μ23051526=0.200=20.0%μ1-μ2.

d. the two-sample z test for p1-p23051526=0.200=20.0%p1-p2.

e. none of these.

A random sample of size nwill be selected from a population, and the proportion of those in the sample who have a Facebook page will be calculated. How would the margin of error for a 95%confidence interval be affected if the sample size were increased from 50to 200?

(a) It remains the same.

(b) It is multiplied by 2.

(c) It is multiplied by 4.

(d) It is divided by 2.

(e) It is divided by 4.

Facebook As part of the Pew Internet and American Life Project, researchers conducted two surveys. The first survey asked a random sample of 1060U.S. teens about their use of social media. A second survey posed similar questions to a random sample of 2003U.S. adults. In these two studies, 71.0%of teens and 58.0%of adults used Facebook. Let plocalid="1654194806576" T3051526=0.200=20.0%pT= the true proportion of all U.S. teens who use Facebook and pA3051526=0.200=20%pA = the true proportion of all U.S. adults who use Facebook. Calculate and interpret a 99%confidence interval for the difference in the true proportions of U.S. teens and adults who use Facebook.

Music and memory Refer to Exercise 87.

a. Construct and interpret a 99%confidence interval for the true mean difference. If you already defined the parameter and checked conditions in Exercise87, you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise .

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