Mathematics, 19.02.2020 02:46 Nathaliasmiles

Bank of America's Consumer Spending Survey collected data on annualcredit card charges in seven different categories of expenditures:transportation, groceries, dining out, household expenses, homefurnishings, apparel, and entertainment (U. S. AirwaysAttache, December 2003). Using data from a sample of 42 creditcard accounts, assume that each account was used to identify theannual credit card charges for groceries (population 1) and theannual credit card charges for dining out (population 2). Using thedifference data, the sample mean difference was = $850, and the sample standard deviationwas sd = $1,123.
a. Formulate the null abd alternative hypothesis 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 can conclude that the population mean 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?

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Answer from: chloelandry

a) Null hypothesis: $\mu_y- \mu_x = 0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

b) $t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{850 -0}{\frac{1123}{\sqrt{42}}}=4.905$

The next step is calculate the degrees of freedom given by:

df=n-1=42-1=41

Now we can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =2*P(t_{(41)}4.905) =7.6x10^{-6}$

So the p value is lower than any significance level given, so then we can conclude that we can to reject the null hypothesis that the difference between the two mean is equal to 0.

c) The confidence interval is given by:

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

For this case we have a confidence of 1-0.05 = 0.95 so we need 0.05 of the area of the t distribution with 41 df on the tails. So we need 0.025 of the area on each tail, and the critical value would be:

$t_{crit}= 2.02$

And if we find the interval we got:

$850- 2.02* \frac{1123}{\sqrt{42}}=499.96$

$850+ 2.02* \frac{1123}{\sqrt{42}}=1200.03$

We are confident 95% that the difference between the two means is between 499.96 and 1200.03

So we have enough evidence to conclude that one mean is higher than the other one, without conduct another hypothesis test because the confidence interval for the difference of means not contain the value of 0. And for this case the groceries would have a higher mean

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Part a

Let put some notation

x=test value for 1 , y = test value for 2

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x = 0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}$

This value is given $\bar d = 850$

The third step would be calculate the standard deviation for the differences.

This value is given also s_d = 1123

Part b

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{850 -0}{\frac{1123}{\sqrt{42}}}=4.905$