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Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment (US Airways Attache, December 2003 ). Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1$)$ and the annual credit card charges for dining out (population $2 ) .$ Using the difference data, the sample mean difference was $d=\$ 850,$ and the sample standard deviation was $s_{d}=\$ 1123 .$
a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
b. Use a. 05 level of significance. Can you conclude that the population means differ? What is the $p$ -value?
c. Which category, groceries or dining out, has a higher population mean annual credit card charge? What is the point estimate of the difference between the population means? What is the 95$\%$ confidence interval estimate of the difference between the population means?

Problem. 17 8. Each note is that new one is smaller than or equal commute. Each one is equal to me. One is bigger than you, so the degree of freedom is the minimum off. Anyone minus one into my storage is 18 and 14, so it's equal toe food correspondent to critical value with offer. April 2.451 tail so T is equal to 1.761 So the actual reason contain old values. Great around 0.761 So the test statistic is equal to x one bar minus X to bar. So for 97 year zero minus 4 to 000 over square. Note on 8800 square over 19 plus 51 Use your square over 15, which is ableto 3.19 point. So it's a very all the tests statistic in the rectory does not have pointy inject. Three point is smaller and it's bigger than 1.761 So we reject then the hypothesis

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.

Number 16, which is a each note. I think that new one is wonder than or equal to YouTube. Each one is that new one is bigger than you two. So the critical value stated in table four corresponding to probability off open 99. So that is equal to 2.33 So the distance statistic in that if they quinto x one more, one is six to bar minus me one minus for you too over square root off signal only squared often anyone waas signal two squared over. And to which approximately equal toe 0.21. So in the very on the this is it within the rejection reason which is the old values greater than 2.33 so as to point someone is smaller than 2.33 So we fail to reject the null hypothesis. So there is no sufficient evidence to support that claim.

In this table. I have summarized the information given in the question about the study on men's and women's average spending habits on leisure activities. So it's sample sizes are both 20. Average expenditure by men on leisure activities is $178.96 for women and the sample standard deviations air, $75.50 dollars for men and women, respectively. So for part A were asked to calculate the estimate on standard error. So this is equal with standard air is calculated like this and then substituting in the numbers from the table. This comes out to 20.16 So that is the estimate for a standard error. And for B were asked if the difference in men's and women's average spending habits on leisure activities is significant at a significance level of Alfa equals 0.5 So we have Alfa equals 0.5 So the claim that we're testing is that there is some difference in spending habits, so the mean for a man minus the mean for women is not equal to zero. So the no hypothesis is would be that the same. So I mean for man minus I mean for women is equal to zero, so we can use the classic method where we compare a test statistic to a critical value. So first, let's calculate the test statistic that's equal to the difference in the sample averages over the standard error, which is equal to 178 minus 96 over the standard air, calculated in part B, part A, and this comes out to 4.68 and now for a critical value. So are degrees of freedom are equal to the smaller sample size. Both are 20 so this that's 20 minus one is 19 and a critical value then is given the degrees of freedom of 19 and the significance level of 0.5 spread across two tails. Because this is a two tailed test, we can look this up in the table, and I have it circled here. So 19 degrees of freedom, 0.5 in two tails gives us a critical value of plus or minus 2.93 and thats plus or minus 2.93 Since this is in a two tailed test. So basically, if we have a a test statistic greater than 2.93 or less than minus 2.93 So now the final step is to compare the test statistic to the critical value and test statistic is quite a bit bigger than the critical value, which means that we reject the no hypothesis and we can say that there is evidence to suggest that there is a difference between the average expenditure by men and women on leisure activities.