Question

an experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. the mean tip percentages and standard deviations are given in the accompanying table along with the sample sizes. 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. complete parts (a) and (b).

the test statistic, ( t ), is ( -4.23 ) (round to two decimal places as needed)
the p - value is ( 0.000 ) (round to three decimal places as needed.)
state the conclusion for the test
a. fail to reject the null hypothesis. there is not sufficient evidence to support the claim that giving candy does result in greater tips
b. reject the null hypothesis. there is not sufficient evidence to support the claim that giving candy does result in greater tips
c. fail to reject the null hypothesis. there is sufficient evidence to support the claim that giving candy does result in greater tips
d. reject the null hypothesis. there is sufficient evidence to support the claim that giving candy does result in greater tips
b. construct the confidence interval suitable for testing the claim in part (a)
( square<mu_1 - mu_2<square )
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the degrees of freedom

For two - sample \(t\) - test with unequal variances, the formula for degrees of freedom (\(df\)) is a bit complex. But for the confidence interval, we can use the conservative approach. Since \(n_1 = n_2=21\), we can also use the formula \(df=\min(n_1 - 1,n_2 - 1)\). Here \(n_1=n_2 = 21\), so \(df=20\).

Step2: Find the critical value

For a one - tailed test (claim \(H_1:\mu_1<\mu_2\) which is equivalent to \(\mu_1-\mu_2 < 0\) when the null hypothesis \(H_0:\mu_1-\mu_2=0\)), if we assume a significance level \(\alpha = 0.05\). Using the \(t\) - distribution table or a calculator, the critical value \(t_{\alpha,df}\) for \(df = 20\) and \(\alpha=0.05\) (one - tailed) is \(t_{0.05,20}=1.725\).

Step3: Calculate the confidence interval

The formula for the confidence interval for \(\mu_1-\mu_2\) when \(\sigma_1
eq\sigma_2\) is \((\bar{x}_1-\bar{x}_2)-t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\)

We have \(\bar{x}_1 = 19.15\), \(\bar{x}_2=21.82\), \(s_1 = 1.33\), \(s_2=2.57\), \(n_1=n_2 = 21\), and for a one - tailed test with \(\alpha = 0.05\), the critical value \(t\) (using the fact that the confidence level for a one - tailed test is \(1 - 2\alpha\) in the formula of the confidence interval) is \(t = 1.725\)

First, calculate \(\bar{x}_1-\bar{x}_2=19.15 - 21.82=- 2.67\)

Then calculate \(\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=\sqrt{\frac{1.33^{2}}{21}+\frac{2.57^{2}}{21}}=\sqrt{\frac{1.7689 + 6.6049}{21}}=\sqrt{\frac{8.3738}{21}}\approx\sqrt{0.3988}\approx0.631\)

The margin of error \(E=t\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=1.725\times0.631\approx1.09\)

The confidence interval is \((-2.67-1.09)<\mu_1-\mu_2<(-2.67 + 1.09)\)

Answer:

\(-3.76<\mu_1-\mu_2<-1.58\)