Before changes to its management, staff and automobile assembly line operation had a scheduled meeting completion time of 15 minutes. Standard deviation of completion time was 1.3 minutes and analyze at the company suspects that under new management the main completion time is now less than 15 minutes to test this claim a random sample of 42 completion times under new management was taken by the analyst the sample had a mean of 14.6 minutes assume that the population is normally distributed can we support at the 0.05 level of significance the claim at the population mean completion time under new management is less than 15 minutes , assume that the population standard deviation of completion times has not changed on the new management perform a one tailed test state the null hypothesis H0 and the alternative hypothesis H1. Determine the type of test statistic to use find the value of the test statistic round to three or more decimal places. Find the P value ran to three or more decimal places.

To test the claim that the population mean completion time under new management is less than 15 minutes, we'll perform a one-tailed hypothesis test using the following steps:

### Step 1: State the Hypotheses
- **Null Hypothesis (\(H_0\))**: The population mean completion time under new management is 15 minutes. \[ H_0: \mu = 15 \]
- **Alternative Hypothesis (\(H_1\))**: The population mean completion time under new management is less than 15 minutes. \[ H_1: \mu < 15 \]

### Step 2: Identify the Type of Test Statistic
Since the population standard deviation is known (1.3 minutes), and the sample size is large (\(n = 42\)), we will use a **z-test**.

### Step 3: Calculate the Test Statistic
The test statistic for a z-test is given by:

\[
z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \(\bar{x} = 14.6\) (sample mean)
- \(\mu_0 = 15\) (population mean under the null hypothesis)
- \(\sigma = 1.3\) (population standard deviation)
- \(n = 42\) (sample size)

Let's plug in the values:

\[
z = \frac{14.6 - 15}{\frac{1.3}{\sqrt{42}}}
\]

I'll calculate this now.The calculated test statistic is approximately \( z = -1.994 \).

### Step 4: Find the P-value
The P-value for a one-tailed z-test can be found by looking up the cumulative probability corresponding to the calculated z-value in the standard normal distribution table.

Since this is a left-tailed test, we are interested in \( P(Z < -1.994) \).

Let's calculate the P-value.The P-value is approximately \( 0.023 \).

### Step 5: Conclusion
Compare the P-value to the significance level \(\alpha = 0.05\):

- If \( \text{P-value} < \alpha \), reject the null hypothesis \(H_0\).
- If \( \text{P-value} \geq \alpha \), do not reject the null hypothesis.

In this case, \( \text{P-value} = 0.023 \) is less than \( \alpha = 0.05 \), so we reject the null hypothesis.

### Conclusion
At the 0.05 level of significance, there is sufficient evidence to support the claim that the population mean completion time under new management is less than 15 minutes.

Would you like more details or have any questions?

### Relative Questions:
1. What is a one-tailed test, and when is it used?
2. How does the P-value help in hypothesis testing?
3. What assumptions must be met to use a z-test?
4. How would the test change if the sample size were smaller?
5. What is the importance of the significance level in hypothesis testing?
6. How do you interpret a rejection of the null hypothesis?
7. What would happen if the population standard deviation were unknown?
8. Can the results of this test be generalized to all such situations?

### Tip:
Always ensure that the conditions for using a specific statistical test are met before applying it.

Learn how to perform a one-tailed hypothesis test using a z-test. This example evaluates whether the mean completion time under new management is less than 15 minutes based on a sample mean of 14.6 minutes and a known population standard deviation of 1.3 minutes. Understand the steps involved, including formulating hypotheses, calculating the test statistic, finding the P-value, and interpreting the results.

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