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You wish to test the following claim (Ha) at a significance level of α=0.005.

      Ho:μ=77.1
      Ha:μ>77.1

You believe the population is normally distributed and **you know the standard deviation** is σ=7.1. You obtain a sample mean of M=78.5 for a sample of size n=42.

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

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value = 

The p-value is...

less than (or equal to) α

greater than α



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 77.1.

There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 77.1.

The sample data support the claim that the population mean is greater than 77.1.

There is not sufficient sample evidence to support the claim that the population mean is greater than 77.1.

To solve this hypothesis testing problem, we will calculate the test statistic and the p-value.

### Given Information:
- Null Hypothesis (H0): μ = 77.1
- Alternative Hypothesis (Ha): μ > 77.1
- Population standard deviation (σ): 7.1
- Sample mean (M): 78.5
- Sample size (n): 42
- Significance level (α): 0.005

### Step 1: Calculate the Test Statistic

The test statistic for a one-sample z-test is calculated using the formula:

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

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

Substituting the values:

\[
z = \frac{78.5 - 77.1}{\frac{7.1}{\sqrt{42}}}
\]

Calculating the denominator:

\[
\frac{7.1}{\sqrt{42}} \approx \frac{7.1}{6.4807} \approx 1.0953
\]

Now, substituting back into the formula:

\[
z = \frac{1.4}{1.0953} \approx 1.278
\]

Thus, the test statistic is:

\[
\text{test statistic} = 1.278 \quad (\text{accurate to three decimal places: } 1.278)
\]

### Step 2: Calculate the p-value

To find the p-value for a one-tailed test, we will use the z-table or a normal distribution calculator. The p-value is the probability of observing a test statistic as extreme as 1.278 or more extreme in the direction of the alternative hypothesis.

Using a z-table or calculator, we find:

\[
\text{p-value} = P(Z > 1.278) \approx 0.1003 \quad (\text{accurate to four decimal places: } 0.1003)
\]

### Step 3: Compare the p-value to α

- The p-value (0.1003) is greater than α (0.005).

### Step 4: Decision

Since the p-value is greater than α, we fail to reject the null hypothesis.

### Step 5: Conclusion

There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 77.1.

### Summary of Results:

1. **Test Statistic**: 1.278
2. **p-value**: 0.1003
3. **Decision**: Fail to reject the null hypothesis.
4. **Conclusion**: There is not sufficient evidence to support the claim that the population mean is greater than 77.1.

### Questions for Further Consideration:

1. What are the implications of failing to reject the null hypothesis in this context?
2. How would the conclusion change if the sample mean were significantly higher?
3. What assumptions are necessary for conducting this z-test?
4. How could the sample size affect the results of this hypothesis test?
5. What is the power of this test given the current sample size and effect size?

### Tip:
Always check the assumptions of normality and independence before conducting hypothesis tests.

You wish to test the following claim (Ha) at a significance level of α=0.005. Ho:μ=77.1, Ha:μ>77.1. You believe the population is normally distributed and you know the standard deviation is σ=7.1. You obtain a sample mean of M=78.5 for a sample of size n=42. What is the test statistic for this sample? (Report answer accurate to three decimal places.) What is the p-value for this sample? (Report answer accurate to four decimal places.) The p-value is less than (or equal to) α, greater than α. 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 77.1, there is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 77.1.

Explore the hypothesis testing process with a one-sample z-test. This example investigates the population mean with a null hypothesis of μ=77.1, a sample mean of M=78.5, and a significance level of α=0.005. Learn to calculate the test statistic and p-value.

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