investigate, a random sample of 27 sales representatives reveals that the mean number of calls made last week was 40. The standard deviation of the sample is 6.1 calls. Using the 0.025 significance level, can we conclude that the mean number of calls per salesperson per week is more than 39? Ho: Hs 39 H1: H> 39 a. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic b. What is your decision regarding Ho? v H0. The mean number of calls is than 39 per week.

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### Hypothesis Testing for the Mean Number of Calls Per Salesperson
To investigate, a random sample of 27 sales representatives reveals that the mean number of calls made last week was 40. The standard deviation of the sample is 6.1 calls. Using the 0.025 significance level, can we conclude that the mean number of calls per salesperson per week is more than 39?

#### Null and Alternative Hypotheses
**H₀:** μ ≤ 39  
**H₁:** μ > 39

#### Questions:

**a. Compute the value of the test statistic. (Round your answer to 3 decimal places.)**

\[ \text{Value of the test statistic} \]

**b. What is your decision regarding \( H_0 \)?**

\[ H_0 \text{. The mean number of calls is} \]  
\[ \text{than 39 per week.} \]

**Explanation of computations and decisions:**

**Step-by-step Guide:**

**1. Compute the Test Statistic:**
To test the hypotheses, we will use a t-test for the sample mean.
The formula for the t-test statistic is:

\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

Where:
- \( \bar{x} \) is the sample mean
- \( \mu_0 \) is the population mean under the null hypothesis
- \( s \) is the sample standard deviation
- \( n \) is the sample size

Given:
- \( \bar{x} = 40 \)
- \( \mu_0 = 39 \)
- \( s = 6.1 \)
- \( n = 27 \)

\[ t = \frac{40 - 39}{6.1 / \sqrt{27}} \]

**2. Calculate the degrees of freedom:**
The degrees of freedom (df) for a t-test for one sample mean is:

\[ \text{df} = n - 1 \]

For this example:

\[ \text{df} = 27 - 1 = 26 \]

**3. Make the Decision:**
Given the 0.025 significance level, we will find the critical value from the t-distribution table for a one-tailed test with 26 df. Compare the test statistic calculated in step 1 to this critical value to decide whether to reject
Transcribed Image Text:### Hypothesis Testing for the Mean Number of Calls Per Salesperson To investigate, a random sample of 27 sales representatives reveals that the mean number of calls made last week was 40. The standard deviation of the sample is 6.1 calls. Using the 0.025 significance level, can we conclude that the mean number of calls per salesperson per week is more than 39? #### Null and Alternative Hypotheses **H₀:** μ ≤ 39 **H₁:** μ > 39 #### Questions: **a. Compute the value of the test statistic. (Round your answer to 3 decimal places.)** \[ \text{Value of the test statistic} \] **b. What is your decision regarding \( H_0 \)?** \[ H_0 \text{. The mean number of calls is} \] \[ \text{than 39 per week.} \] **Explanation of computations and decisions:** **Step-by-step Guide:** **1. Compute the Test Statistic:** To test the hypotheses, we will use a t-test for the sample mean. The formula for the t-test statistic is: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where: - \( \bar{x} \) is the sample mean - \( \mu_0 \) is the population mean under the null hypothesis - \( s \) is the sample standard deviation - \( n \) is the sample size Given: - \( \bar{x} = 40 \) - \( \mu_0 = 39 \) - \( s = 6.1 \) - \( n = 27 \) \[ t = \frac{40 - 39}{6.1 / \sqrt{27}} \] **2. Calculate the degrees of freedom:** The degrees of freedom (df) for a t-test for one sample mean is: \[ \text{df} = n - 1 \] For this example: \[ \text{df} = 27 - 1 = 26 \] **3. Make the Decision:** Given the 0.025 significance level, we will find the critical value from the t-distribution table for a one-tailed test with 26 df. Compare the test statistic calculated in step 1 to this critical value to decide whether to reject
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