You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002.

      Ho:μ=62.7Ho:μ=62.7
      Ha:μ>62.7Ha:μ>62.7

You believe the population is normally distributed and you know the standard deviation is σ=16.5σ=16.5. You obtain a sample mean of M=67.6M=67.6 for a sample of size n=71n=71.

What is the critical value for this test? (Report answer accurate to three decimal places.)
critical value = 

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic = 

The test statistic is...

in the critical region

not in the critical region


This test statistic leads to a decision to...

reject the null

accept the null

fail to reject the null


As such, the final conclusion is that...

There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 62.7.

There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 62.7.

The sample data support the claim that the population mean is greater than 62.7.

There is not sufficient sample evidence to support the claim that the population mean is greater than 62.7.

Let's go step by step to solve the problem.

### 1. Determine the critical value:
Since this is a one-tailed test (specifically a right-tailed test because \( H_a \) is \( \mu > 62.7 \)), we need to find the critical value corresponding to a significance level \( \alpha = 0.002 \).

Using the standard normal distribution table, the critical value \( z_{\alpha} \) for \( \alpha = 0.002 \) is:

\[
\text{Critical value} = 2.878
\]

### 2. Calculate the test statistic:
The formula for the test statistic \( z \) is:

\[
z = \frac{M - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \( M = 67.6 \) (sample mean)
- \( \mu = 62.7 \) (population mean under the null hypothesis)
- \( \sigma = 16.5 \) (population standard deviation)
- \( n = 71 \) (sample size)

Now, substitute the values into the formula:

\[
z = \frac{67.6 - 62.7}{\frac{16.5}{\sqrt{71}}}
\]

Calculate the denominator:

\[
\frac{16.5}{\sqrt{71}} \approx \frac{16.5}{8.4261} \approx 1.9585
\]

Now, calculate the test statistic:

\[
z \approx \frac{4.9}{1.9585} \approx 2.502
\]

### 3. Compare the test statistic with the critical value:
- The test statistic is \( z \approx 2.502 \).
- The critical value for this test is \( z_{\alpha} = 2.878 \).

Since \( 2.502 \) is **not greater than** \( 2.878 \), the test statistic is **not in the critical region**.

### 4. Make a decision:
Since the test statistic is **not in the critical region**, we **fail to reject the null hypothesis**.

### 5. Final conclusion:
There is **not sufficient sample evidence to support** the claim that the population mean is greater than 62.7.

If you have any further questions or need more details, feel free to ask.

Here are 8 related questions you might be interested in:
1. How would the decision change if the significance level were 0.05 instead of 0.002?
2. What is the p-value associated with the test statistic of 2.502?
3. How does the sample size influence the test statistic?
4. What would happen if the population standard deviation were smaller?
5. How would the test statistic change if the sample mean were lower?
6. What are the implications of using a higher or lower significance level (\( \alpha \))?
7. Can you explain the difference between rejecting and failing to reject the null hypothesis?
8. How would the result differ if this were a two-tailed test instead of a one-tailed test?

**Tip:** When working with hypothesis tests, always carefully distinguish between one-tailed and two-tailed tests, as this impacts the critical value and the final decision.

Learn how to conduct hypothesis testing for a population mean with a known standard deviation. This problem involves testing a claim at a specific significance level using a one-tailed z-test. Calculate critical values, test statistics, and make decisions based on sample data.

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