Answer :
Let's go through the steps to address the hypothesis test about the service ratings for the two consultants.
### a. State the null and alternative hypotheses.
We are testing whether the consultant with more experience (Consultant A) has a higher mean service rating than the consultant with less experience (Consultant B).
- Null Hypothesis (H₀):
The mean service rating for Consultant A is less than or equal to the mean service rating for Consultant B.
[tex]\[ H_0: \mu_1 - \mu_2 \leq 0 \][/tex]
- Alternative Hypothesis (Hₐ):
The mean service rating for Consultant A is greater than the mean service rating for Consultant B.
[tex]\[ H_a: \mu_1 - \mu_2 > 0 \][/tex]
### b. Compute the value of the test statistic.
To compute the value of the test statistic, we follow these steps:
#### Step 1: Calculate the pooled standard deviation
The pooled standard deviation is a combined estimate of the common standard deviation for both groups. It integrates the variances and sample sizes of both groups:
[tex]\[ \text{Pooled Standard Deviation (pooled_std)} = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \][/tex]
Given:
- [tex]\( n_1 = 16 \)[/tex]
- [tex]\( s_1 = 0.64 \)[/tex]
- [tex]\( n_2 = 10 \)[/tex]
- [tex]\( s_2 = 0.75 \)[/tex]
Thus:
[tex]\[ \text{pooled_std} = 0.6833282520136278 \][/tex]
#### Step 2: Calculate the standard error of the difference in means
The standard error of the difference in means is:
[tex]\[ \text{SE}_{\text{diff}} = \text{pooled_std} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \][/tex]
Given:
[tex]\[ \text{SE}_{\text{diff}} = 0.27545842472140875 \][/tex]
#### Step 3: Compute the value of the test statistic (t-statistic)
The t-statistic is calculated using the formula:
[tex]\[ t = \frac{\overline{x}_1 - \overline{x}_2}{\text{SE}_{\text{diff}}} \][/tex]
Given:
- [tex]\( \overline{x}_1 = 6.82 \)[/tex]
- [tex]\( \overline{x}_2 = 6.25 \)[/tex]
- [tex]\(\text{SE}_{\text{diff}}= 0.27545842472140875\)[/tex]
Thus:
[tex]\[ t = \frac{6.82 - 6.25}{0.27545842472140875} = 2.0692777887497287 \approx 2.07 \text{ (to 2 decimals)} \][/tex]
#### Step 4: Compute the degrees of freedom (df)
The degrees of freedom for a two-sample t-test is given by:
[tex]\[ df = n_1 + n_2 - 2 \][/tex]
Given:
- [tex]\( n_1 = 16 \)[/tex]
- [tex]\( n_2 = 10 \)[/tex]
Thus:
[tex]\[ df = 16 + 10 - 2 \][/tex]
[tex]\[ df = 24 \][/tex]
### Summary
a. State the hypotheses:
[tex]\[ H_0: \mu_1 - \mu_2 \leq 0 \][/tex]
[tex]\[ H_a: \mu_1 - \mu_2 > 0 \][/tex]
b. Compute the test statistic:
[tex]\[ t = 2.07 \][/tex]
Degrees of freedom:
[tex]\[ df = 24 \][/tex]
The critical value for a one-tailed test at [tex]\(\alpha = 0.05\)[/tex] with 24 degrees of freedom can be obtained from the t-distribution table or using statistical software. If the computed t-statistic is greater than the critical t-value, we reject [tex]\(H_0\)[/tex] and conclude that Consultant A has a significantly higher mean service rating than Consultant B.
### a. State the null and alternative hypotheses.
We are testing whether the consultant with more experience (Consultant A) has a higher mean service rating than the consultant with less experience (Consultant B).
- Null Hypothesis (H₀):
The mean service rating for Consultant A is less than or equal to the mean service rating for Consultant B.
[tex]\[ H_0: \mu_1 - \mu_2 \leq 0 \][/tex]
- Alternative Hypothesis (Hₐ):
The mean service rating for Consultant A is greater than the mean service rating for Consultant B.
[tex]\[ H_a: \mu_1 - \mu_2 > 0 \][/tex]
### b. Compute the value of the test statistic.
To compute the value of the test statistic, we follow these steps:
#### Step 1: Calculate the pooled standard deviation
The pooled standard deviation is a combined estimate of the common standard deviation for both groups. It integrates the variances and sample sizes of both groups:
[tex]\[ \text{Pooled Standard Deviation (pooled_std)} = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \][/tex]
Given:
- [tex]\( n_1 = 16 \)[/tex]
- [tex]\( s_1 = 0.64 \)[/tex]
- [tex]\( n_2 = 10 \)[/tex]
- [tex]\( s_2 = 0.75 \)[/tex]
Thus:
[tex]\[ \text{pooled_std} = 0.6833282520136278 \][/tex]
#### Step 2: Calculate the standard error of the difference in means
The standard error of the difference in means is:
[tex]\[ \text{SE}_{\text{diff}} = \text{pooled_std} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \][/tex]
Given:
[tex]\[ \text{SE}_{\text{diff}} = 0.27545842472140875 \][/tex]
#### Step 3: Compute the value of the test statistic (t-statistic)
The t-statistic is calculated using the formula:
[tex]\[ t = \frac{\overline{x}_1 - \overline{x}_2}{\text{SE}_{\text{diff}}} \][/tex]
Given:
- [tex]\( \overline{x}_1 = 6.82 \)[/tex]
- [tex]\( \overline{x}_2 = 6.25 \)[/tex]
- [tex]\(\text{SE}_{\text{diff}}= 0.27545842472140875\)[/tex]
Thus:
[tex]\[ t = \frac{6.82 - 6.25}{0.27545842472140875} = 2.0692777887497287 \approx 2.07 \text{ (to 2 decimals)} \][/tex]
#### Step 4: Compute the degrees of freedom (df)
The degrees of freedom for a two-sample t-test is given by:
[tex]\[ df = n_1 + n_2 - 2 \][/tex]
Given:
- [tex]\( n_1 = 16 \)[/tex]
- [tex]\( n_2 = 10 \)[/tex]
Thus:
[tex]\[ df = 16 + 10 - 2 \][/tex]
[tex]\[ df = 24 \][/tex]
### Summary
a. State the hypotheses:
[tex]\[ H_0: \mu_1 - \mu_2 \leq 0 \][/tex]
[tex]\[ H_a: \mu_1 - \mu_2 > 0 \][/tex]
b. Compute the test statistic:
[tex]\[ t = 2.07 \][/tex]
Degrees of freedom:
[tex]\[ df = 24 \][/tex]
The critical value for a one-tailed test at [tex]\(\alpha = 0.05\)[/tex] with 24 degrees of freedom can be obtained from the t-distribution table or using statistical software. If the computed t-statistic is greater than the critical t-value, we reject [tex]\(H_0\)[/tex] and conclude that Consultant A has a significantly higher mean service rating than Consultant B.
