We want to conduct a hypothesis test of the claim that the population mean score on a nationwide examination in anthropology is different from 519. So, we choose a random sample of exam scores. The sample has a mean of  499 and a standard deviation of 76.

For each of the following sampling scenarios, choose an appropriate test statistic for our hypothesis test on the population mean. Then calculate that statistic. Round your answers to two decimal places. (a)The sample has size 19, and it is from a normally distributed population with a known standard deviation of 76. (b)The sample has size 13, and it is from a normally distributed population with an unknown standard deviation.


To conduct the hypothesis test for each scenario, we need to determine the appropriate test statistic based on the sample size and whether the population standard deviation is known or unknown. Here's how to proceed for each case:

### Given:
- Population mean under the null hypothesis (\(\mu_0\)): 519
- Sample mean (\(\bar{x}\)): 499
- Population standard deviation (\(\sigma\)): 76 (known for scenario (a))
- Sample standard deviation (\(s\)): 76 (given for scenario (b))
- Significance level (\(\alpha\)): Not specified (typically \(0.05\) is assumed unless otherwise stated).

### (a) **Sample size 19 and known population standard deviation:**

#### 1. Determine the test statistic:
Since the population standard deviation is known and the sample size is 19 (less than 30), we use the **z-test** statistic. 

#### 2. Calculate the z-test statistic:
The z-test statistic is calculated using the formula:
\[
z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]

Plugging in the values:
\[
z = \frac{499 - 519}{\frac{76}{\sqrt{19}}} = \frac{-20}{\frac{76}{4.359}} = \frac{-20}{17.43} \approx -1.15
\]

So, the test statistic \(z\) is approximately **-1.15**.

### (b) **Sample size 13 and unknown population standard deviation:**

#### 1. Determine the test statistic:
Since the population standard deviation is unknown and the sample size is 13 (less than 30), we use the **t-test** statistic.

#### 2. Calculate the t-test statistic:
The t-test statistic is calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]

Plugging in the values:
\[
t = \frac{499 - 519}{\frac{76}{\sqrt{13}}} = \frac{-20}{\frac{76}{3.606}} = \frac{-20}{21.07} \approx -0.95
\]

So, the test statistic \(t\) is approximately **-0.95**.

### Summary of Results:
- **(a)** For the sample size of 19 with a known population standard deviation, the test statistic \(z\) is **-1.15**.
- **(b)** For the sample size of 13 with an unknown population standard deviation, the test statistic \(t\) is **-0.95**.

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Would you like further details or have any questions?

### Related Questions:

1. How do you interpret the results of the z-test and t-test?
2. What are the critical z-values for a significance level of 0.05 in a two-tailed test?
3. How would the conclusion change if the test statistic was positive instead of negative?
4. What is the difference between the z-test and the t-test?
5. How does sample size affect the choice between a z-test and a t-test?
6. What assumptions are required to use the z-test and t-test?
7. How can you calculate the p-value for each test statistic?
8. What are Type I and Type II errors in the context of hypothesis testing?

**Tip:** When the population standard deviation is known and the sample size is small, the z-test is still appropriate, but when it's unknown, the t-test is used.

Explore hypothesis testing for the claim that the population mean score on a nationwide examination in anthropology differs from 519. Understand how to calculate z-test and t-test statistics with provided sample data (mean of 499, standard deviation of 76) for different sample sizes.

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