A random sample of 78 eighth grade​ students' scores on a national mathematics assessment test has a mean score of 278. This test result prompts a state school administrator to declare that the mean score for the​ state's eighth graders on this exam is more than 270. Assume that the population standard deviation is 40. At alphaequals0.11​, is there enough evidence to support the​ administrator's claim? Complete parts​ (a) through​ (e). ​(c) Find the​ P-value.

To solve this hypothesis test and find the P-value, we need to follow these steps:

### Given:
- Sample size (\(n\)) = 78
- Sample mean (\(\bar{x}\)) = 278
- Population standard deviation (\(\sigma\)) = 40
- Hypothesized population mean (\(\mu_0\)) = 270
- Significance level (\(\alpha\)) = 0.11

### Step 1: Set up the hypotheses.
- Null hypothesis (\(H_0\)): \(\mu = 270\) (The mean score is 270)
- Alternative hypothesis (\(H_a\)): \(\mu > 270\) (The mean score is greater than 270)

### Step 2: Calculate the test statistic.
We use the Z-test for the population mean because the population standard deviation is known. The formula for the Z-test statistic is:

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

Where:
- \(\bar{x}\) is the sample mean,
- \(\mu_0\) is the hypothesized population mean,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.

Substitute the given values into the formula:

\[
Z = \frac{278 - 270}{\frac{40}{\sqrt{78}}}
\]

First, calculate the denominator:

\[
\frac{40}{\sqrt{78}} = \frac{40}{8.83} \approx 4.53
\]

Now, calculate the Z-score:

\[
Z = \frac{278 - 270}{4.53} \approx \frac{8}{4.53} \approx 1.77
\]

### Step 3: Find the P-value.
Since this is a one-tailed test (right-tailed), we can find the P-value by looking up the Z-value in the standard normal distribution table, or using a calculator. The Z-value of 1.77 corresponds to a P-value of approximately **0.0384**.

### Step 4: Conclusion.
Now that we have the P-value, we can compare it to the significance level \(\alpha = 0.11\).

- If the P-value is less than \(\alpha\), we reject the null hypothesis.
- If the P-value is greater than \(\alpha\), we fail to reject the null hypothesis.

In this case, the P-value (0.0384) is less than the significance level (\(\alpha = 0.11\)). Therefore, we reject the null hypothesis.

### Final Answer:
There is enough evidence to support the administrator's claim that the mean score for the state's eighth graders is greater than 270.

---

Would you like a more detailed explanation of any part of the process or further assistance with related concepts?

### Here are some related questions for you to explore:
1. How do you perform a Z-test for proportions?
2. What is the difference between a one-tailed and a two-tailed test?
3. How do you calculate the confidence interval for a population mean using the Z-distribution?
4. What are the assumptions for using a Z-test?
5. How does increasing the sample size affect the results of a hypothesis test?

**Tip**: Always ensure that your sample size is large enough (typically \(n > 30\)) for the Z-test to be valid when the population standard deviation is known.

A random sample of 78 eighth grade students' scores on a national mathematics assessment test has a mean score of 278. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 270. Assume that the population standard deviation is 40. At alpha = 0.11, is there enough evidence to support the administrator's claim? Complete parts (a) through (e). (c) Find the P-value.

Explore the process of hypothesis testing with a Z-test for a sample of 78 eighth grade students' mathematics assessment scores. Learn how to calculate the P-value and test the administrator's claim that the mean score is greater than 270. This problem involves key concepts such as Z-scores, normal distribution, and hypothesis testing at a significance level of 0.11.

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