Answer :
### Step-by-Step Solution:
Let's go through the solution of this hypothesis test step-by-step:
#### Given Data:
- No candy group:
- [tex]\( n_1 = 33 \)[/tex]
- Sample mean [tex]\( \bar{x}_1 = 1936 \)[/tex]
- Sample standard deviation [tex]\( s_1 = 154 \)[/tex]
- Two candies group:
- [tex]\( n_2 = 33 \)[/tex]
- Sample mean [tex]\( \bar{x}_2 = 2216 \)[/tex]
- Sample standard deviation [tex]\( s_2 = 257 \)[/tex]
#### Part (a): Hypotheses
To determine whether giving candy results in greater tips, we need to set up the null and alternative hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no difference in mean tips between the two groups (i.e., the means are equal). This can be stated as:
[tex]\[ H_0: \mu_1 = \mu_2 \][/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean tips are greater for the group given candy compared to the group not given candy. This can be stated as:
[tex]\[ H_1: \mu_1 < \mu_2 \][/tex]
Given the multiple-choice options:
Option (D) correctly states the hypotheses:
- [tex]\( H_0: \mu_1 = \mu_2 \)[/tex]
- [tex]\( H_1: \mu_1 < \mu_2 \)[/tex]
#### Part (b): Test Statistic
We will use a t-test for the difference in means assuming unequal variances (Welch’s t-test).
The test statistic [tex]\( t \)[/tex] is calculated as follows:
[tex]\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]
Using the given data:
[tex]\[ t = \frac{1936 - 2216}{\sqrt{\frac{154^2}{33} + \frac{257^2}{33}}} \][/tex]
Substitute the values:
[tex]\[ t \approx \frac{-280}{\sqrt{\frac{23716}{33} + \frac{66049}{33}}} \][/tex]
[tex]\[ t \approx \frac{-280}{\sqrt{719.27 + 2001.48}} \][/tex]
[tex]\[ t \approx \frac{-280}{\sqrt{2720.75}} \][/tex]
[tex]\[ t \approx \frac{-280}{52.15} \][/tex]
[tex]\[ t \approx -5.369 \][/tex]
The test statistic, rounded to three decimal places, is:
[tex]\[ t \approx -5.369 \][/tex]
So, the test statistic [tex]\( t \)[/tex] is [tex]\( -5.369 \)[/tex].
#### Conclusion
Using the 0.001 significance level, we compare the p-value to determine whether to reject the null hypothesis.
The p-value associated with the test statistic [tex]\( t \approx -5.369 \)[/tex] is extremely small (approximately [tex]\( 9.216 \times 10^{-7} \)[/tex]), which is much less than 0.001.
Since the p-value is less than the significance level of 0.001, we reject the null hypothesis. This provides strong evidence to support the claim that giving candy to dining parties results in greater tips.
Thus, the answer to the question regarding the test statistic is:
[tex]\[ t \approx -5.369 \][/tex]
Let's go through the solution of this hypothesis test step-by-step:
#### Given Data:
- No candy group:
- [tex]\( n_1 = 33 \)[/tex]
- Sample mean [tex]\( \bar{x}_1 = 1936 \)[/tex]
- Sample standard deviation [tex]\( s_1 = 154 \)[/tex]
- Two candies group:
- [tex]\( n_2 = 33 \)[/tex]
- Sample mean [tex]\( \bar{x}_2 = 2216 \)[/tex]
- Sample standard deviation [tex]\( s_2 = 257 \)[/tex]
#### Part (a): Hypotheses
To determine whether giving candy results in greater tips, we need to set up the null and alternative hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no difference in mean tips between the two groups (i.e., the means are equal). This can be stated as:
[tex]\[ H_0: \mu_1 = \mu_2 \][/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean tips are greater for the group given candy compared to the group not given candy. This can be stated as:
[tex]\[ H_1: \mu_1 < \mu_2 \][/tex]
Given the multiple-choice options:
Option (D) correctly states the hypotheses:
- [tex]\( H_0: \mu_1 = \mu_2 \)[/tex]
- [tex]\( H_1: \mu_1 < \mu_2 \)[/tex]
#### Part (b): Test Statistic
We will use a t-test for the difference in means assuming unequal variances (Welch’s t-test).
The test statistic [tex]\( t \)[/tex] is calculated as follows:
[tex]\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]
Using the given data:
[tex]\[ t = \frac{1936 - 2216}{\sqrt{\frac{154^2}{33} + \frac{257^2}{33}}} \][/tex]
Substitute the values:
[tex]\[ t \approx \frac{-280}{\sqrt{\frac{23716}{33} + \frac{66049}{33}}} \][/tex]
[tex]\[ t \approx \frac{-280}{\sqrt{719.27 + 2001.48}} \][/tex]
[tex]\[ t \approx \frac{-280}{\sqrt{2720.75}} \][/tex]
[tex]\[ t \approx \frac{-280}{52.15} \][/tex]
[tex]\[ t \approx -5.369 \][/tex]
The test statistic, rounded to three decimal places, is:
[tex]\[ t \approx -5.369 \][/tex]
So, the test statistic [tex]\( t \)[/tex] is [tex]\( -5.369 \)[/tex].
#### Conclusion
Using the 0.001 significance level, we compare the p-value to determine whether to reject the null hypothesis.
The p-value associated with the test statistic [tex]\( t \approx -5.369 \)[/tex] is extremely small (approximately [tex]\( 9.216 \times 10^{-7} \)[/tex]), which is much less than 0.001.
Since the p-value is less than the significance level of 0.001, we reject the null hypothesis. This provides strong evidence to support the claim that giving candy to dining parties results in greater tips.
Thus, the answer to the question regarding the test statistic is:
[tex]\[ t \approx -5.369 \][/tex]
