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).

state the conclusion for the test.

a. reject the null hypothesis. there is not sufficient evidence to support the claim that giving candy does result in greater tips.

b. fall to reject the null hypothesis. there is sufficient evidence to support the claim that giving candy does result in greater tips.

c. reject the null hypothesis. there is sufficient evidence to support the claim that giving candy does result in greater tips.

d. fall to reject the null hypothesis. there is

b. construct the confidence interval suitable for testing the claim that giving candy does result in greater tips. (round to two decimal places as needed.)

-3.06 < μ₁ - μ₂ < -1.49

does the confidence interval support the conclusion found with the hypothesis test?

yes, because the confidence interval contains only negative values. zero. only positive values.

Explanation:

Step1: Determine the null and alternative hypotheses

The claim is that giving candy results in greater tips. Let \(\mu_1\) be the mean tip percentage for no - candy and \(\mu_2\) be the mean tip percentage for two - candies. The null hypothesis \(H_0:\mu_1=\mu_2\) (no difference in mean tip percentages) and the alternative hypothesis \(H_1:\mu_1<\mu_2\) (mean tip percentage for no - candy is less than for two - candies)

Step2: Analyze the p - value

Given \(p - value = 0.000\). The decision rule for hypothesis testing is: If \(p - value<\alpha\) (commonly \(\alpha = 0.05\)), we reject the null hypothesis. Since \(0.000<0.05\)

Step3: State the conclusion

We reject the null hypothesis. There is sufficient evidence to support the claim that giving candy (in this case, two candies) results in greater tips.

Step4: Construct the confidence interval

The formula for the confidence interval for \(\mu_1-\mu_2\) (when variances are equal) is \((\bar{x}_1-\bar{x}_2)-t_{\alpha/2}\cdot s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+t_{\alpha/2}\cdot s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)

First, calculate \(s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\)

\(n_1=n_2 = 36\), \(\bar{x}_1=19.24\), \(s_1 = 1.44\), \(\bar{x}_2=21.52\), \(s_2=2.39\)

\(s_p=\sqrt{\frac{(36 - 1)\times(1.44)^2+(36 - 1)\times(2.39)^2}{36+36 - 2}}\)

\(=\sqrt{\frac{35\times2.0736+35\times5.7121}{70}}\)

\(=\sqrt{\frac{72.576+199.9235}{70}}=\sqrt{\frac{272.4995}{70}}\approx\sqrt{3.89285}\approx1.973\)

For a one - tailed test with \(\alpha = 0.05\) and \(df=n_1 + n_2-2=70\), \(t_{\alpha}=1.667\)

\(\bar{x}_1-\bar{x}_2=19.24 - 21.52=- 2.28\)

The confidence interval is \(-2.28-1.667\times1.973\times\sqrt{\frac{1}{36}+\frac{1}{36}}<-2.28 + 1.667\times1.973\times\sqrt{\frac{1}{36}+\frac{1}{36}}\)

\(\sqrt{\frac{1}{36}+\frac{1}{36}}=\sqrt{\frac{2}{36}}=\frac{\sqrt{2}}{6}\approx0.236\)

\(-2.28-1.667\times1.973\times0.236<-2.28 + 1.667\times1.973\times0.236\)

\(1.667\times1.973\times0.236\approx1.667\times0.466\approx0.777\)

The confidence interval is \(-2.28-0.777=-3.06\) to \(-2.28 + 0.777=-1.50\) (rounded to two decimal places)

Since the confidence interval contains only negative values, it supports the conclusion that \(\mu_1<\mu_2\)

Answer:

A. Reject the null hypothesis. There is sufficient evidence to support the claim that giving candy results in greater tips.
b. The confidence interval is \(-3.06<\mu_1-\mu_2<-1.49\) (rounded to two decimal places) and it contains only negative values, so yes, the confidence interval supports the conclusion.