teach me how to do the following in excel: Fry Brothers Heating and Air Conditioning Incorporated employs Gael Clark and Leslie Murnen to make service calls to repair furnaces and air-conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days George Murnen made an average of 5.02 calls per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen.

At the 0.05 significance level, is there a difference in the mean number of calls per day between the two employees?

Required:
State the null and alternate hypotheses.



Compute the test statistic.

Note: Negative values should be indicated by a minus sign. Round your answer to 2 decimal places.



Compute the p-value. NOTE: p-value calculators are available through online search.

Note: Round your answer to 4 decimal places.



What is your decision regarding the null hypothesis?



At the 0.05 significance level, is there a difference in the mean number of calls per day between the two employees?

To determine if there is a significant difference in the mean number of service calls per day between Gael Clark and Leslie Murnen, we need to conduct a two-sample z-test. Below are the steps to perform the hypothesis test in Excel.

### Step 1: State the Hypotheses
- **Null Hypothesis (\(H_0\))**: There is no difference in the mean number of calls per day between the two employees. Mathematically, \(H_0: \mu_1 = \mu_2\).
- **Alternative Hypothesis (\(H_1\))**: There is a difference in the mean number of calls per day between the two employees. Mathematically, \(H_1: \mu_1 \neq \mu_2\).

### Step 2: Calculate the Test Statistic
The formula for the z-test statistic for two independent samples is:

\[
z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}
\]

Where:
- \(\bar{X}_1 = 4.77\) (Mean for Gael Clark)
- \(\bar{X}_2 = 5.02\) (Mean for Leslie Murnen)
- \(\sigma_1 = 1.05\) (Standard deviation for Gael Clark)
- \(\sigma_2 = 1.23\) (Standard deviation for Leslie Murnen)
- \(n_1 = 40\) (Sample size for Gael Clark)
- \(n_2 = 50\) (Sample size for Leslie Murnen)

**In Excel:**
1. Input the values into Excel:
   - \(\bar{X}_1\) (mean1) = 4.77
   - \(\bar{X}_2\) (mean2) = 5.02
   - \(\sigma_1\) (stdev1) = 1.05
   - \(\sigma_2\) (stdev2) = 1.23
   - \(n_1\) (n1) = 40
   - \(n_2\) (n2) = 50

2. Calculate the test statistic (z-value):
   - In a cell, type the formula:
   ```excel
   = (mean1 - mean2) / SQRT((stdev1^2/n1) + (stdev2^2/n2))
   ```
   - Substitute with the actual values:
   ```excel
   = (4.77 - 5.02) / SQRT((1.05^2/40) + (1.23^2/50))
   ```
   - The result will be your z-value.

**Result:** \(z\) = -1.07 (rounded to two decimal places)

### Step 3: Calculate the p-value
The p-value for a two-tailed test can be found using the `NORM.S.DIST` function in Excel.

**In Excel:**
1. Use the formula:
   ```excel
   = 2 * NORM.S.DIST(-ABS(z), TRUE)
   ```
   - Substitute \(z\) with the value obtained from Step 2:
   ```excel
   = 2 * NORM.S.DIST(-ABS(-1.07), TRUE)
   ```
2. The result will be the p-value.

**Result:** p-value ≈ 0.2843 (rounded to four decimal places)

### Step 4: Decision Regarding the Null Hypothesis
- **Significance level (\(\alpha\))**: 0.05
- If the p-value is less than \(\alpha\), reject the null hypothesis. Otherwise, do not reject it.

**Decision:**
- Since the p-value (0.2843) is greater than the significance level (0.05), **we do not reject the null hypothesis**.

### Conclusion
At the 0.05 significance level, there is **no statistically significant difference** in the mean number of calls per day between Gael Clark and Leslie Murnen.

### Would you like more details or have any questions? Here are some related questions you might consider:
1. How would the test change if the standard deviations were unknown and estimated from the sample data?
2. How can you visualize the difference in means using a graph in Excel?
3. What are Type I and Type II errors, and how do they relate to this test?
4. How would the test procedure change if we were conducting a one-tailed test?
5. How would you calculate the confidence interval for the difference in means?

**Tip:** Always check the assumptions of the statistical test you are using, such as the normality of the data distribution and the equality of variances.

Teach me how to do the following in Excel: Fry Brothers Heating and Air Conditioning Incorporated employs Gael Clark and Leslie Murnen to make service calls to repair furnaces and air-conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days George Murnen made an average of 5.02 calls per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. At the 0.05 significance level, is there a difference in the mean number of calls per day between the two employees? Required: State the null and alternate hypotheses. Compute the test statistic. Compute the p-value. What is your decision regarding the null hypothesis? At the 0.05 significance level, is there a difference in the mean number of calls per day between the two employees?

Learn how to conduct a two-sample z-test to compare the mean number of service calls per day between two employees at Fry Brothers Heating and Air Conditioning. This guide includes the step-by-step process, including hypothesis testing, calculating the test statistic, and determining the p-value using Excel. Suitable for undergraduate statistics students.

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