Question

Scores of an IQ test are normally distributed. A sample of 25 IQ scores had standard deviation s= 8. The developer of the test claims that the population standard deviation is greater than =17. Do these data provide sufficient evidence to support this claim. Use a= 0.10 level of significance. First state the appropriate null and alternative hypotheses

Answer 1

[AI-generated answer — Bot GPT-4o]

To determine whether the sample data provide sufficient evidence to support the claim that the population standard deviation is greater than 17 at a 0.10 level of significance, we need to perform a hypothesis test for a single standard deviation.

The appropriate null and alternative hypotheses can be stated as follows:

- Null hypothesis (\(H_0\)): \(\sigma \leq 17\) (The population standard deviation is less than or equal to 17)
- Alternative hypothesis (\(H_1\)): \(\sigma > 17\) (The population standard deviation is greater than 17)

Next, we perform the hypothesis test using the Chi-Square (\(\chi^2\)) distribution.

Given:
- Sample size (\(n\)) = 25
- Sample standard deviation (\(s\)) = 8
- Population standard deviation claim (\(\sigma_0\)) = 17
- Level of significance (\(\alpha\)) = 0.10

The chi-square test statistic for variance can be calculated using the formula:

\[
\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
\]

Plugging in the values:

\[
\chi^2 = \frac{(25-1) \cdot 8^2}{17^2}
\]

\[
\chi^2 = \frac{24 \cdot 64}{289}
\]

\[
\chi^2 = \frac{1536}{289}
\]

\[
\chi^2 \approx 5.31
\]

Now, we need to find the critical value from the chi-square distribution table for \(n - 1 = 24\) degrees of freedom at the 0.10 level of significance (upper tail critical value for a one-tailed test).

Using a chi-square distribution table or a calculator, we find that the critical value (\(\chi^2_{0.10, 24}\)) is approximately 34.382.

Finally, we compare the test statistic to the critical value:

- If \(\chi^2 \geq 34.382\), we reject the null hypothesis.
- If \(\chi^2 < 34.382\), we fail to reject the null hypothesis.

Since \(\chi^2 \approx 5.31\) is much less than 34.382, we fail to reject the null hypothesis.

Therefore, based on the sample data and at the 0.10 level of significance, there is insufficient evidence to support the claim that the population standard deviation is greater than 17.

Answer 2

Find the critical value

Answer 3

[AI-generated answer — Bot GPT-4o]

To find the critical value for the chi-square (\(\chi^2\)) distribution at a 0.10 level of significance with 24 degrees of freedom (\(df = n-1\)), we typically consult a chi-square distribution table or use statistical software.

Here, we are performing a one-tailed test in the upper tail because we are testing if the population standard deviation is greater than a certain value.

### Steps to Obtain the Critical Value (\(\chi^2_{\text{critical}}\)):

1. **Significance Level (\(\alpha\))**: 0.10
2. **Degrees of Freedom (df)**: \(24 = 25 - 1\)

#### Using Chi-Square Distribution Table:
Look up the critical value in the chi-square table for \(df = 24\) at a 0.10 level of significance in the upper tail.

### Chi-Square Critical Value Table Excerpt:
| Degrees of Freedom | 0.10 (Upper Tail) |
|-------------------- |----------------- |
| 24 | 33.196 |

From the table, we find that:

**The critical value (\(\chi^2_{\text{critical}}\)) for \(24\) degrees of freedom at the \(\alpha = 0.10\) level of significance is approximately \(33.196\).**

With this critical value, we can compare it against our calculated test statistic to determine whether to reject the null hypothesis:

- If \(\chi^2 \geq 33.196\), we reject the null hypothesis.
- If \(\chi^2 < 33.196\), we fail to reject the null hypothesis.

### Summary:
- Calculated test statistic: \(\chi^2 \approx 5.31\)
- Critical value: \(\chi^2_{\text{critical}} = 33.196\)

Since \(5.31 < 33.196\), we fail to reject the null hypothesis. Hence, there is insufficient evidence to support the claim that the population standard deviation is greater than 17 at the 0.10 level of significance.

Answer 4

Next compute the test statistic

Answer 5

[AI-generated answer — Bot GPT-4o]

To compute the test statistic for the hypothesis test regarding the population standard deviation, we use the chi-square (\(\chi^2\)) test statistic for variance. The formula we use is:

\[
\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
\]

Given:
- Sample size (\(n\)) = 25
- Sample standard deviation (\(s\)) = 8
- Population standard deviation claim (\(\sigma_0\)) = 17

### Steps to Compute the Test Statistic:

1. **First, compute the sample variance (\(s^2\))**:
\[
s^2 = 8^2 = 64
\]

2. **Next, substitute the values into the chi-square formula**:
\[
\chi^2 = \frac{(25-1) \cdot 64}{17^2}
\]

3. **Calculate the denominator (\(\sigma_0^2\))**:
\[
\sigma_0^2 = 17^2 = 289
\]

4. **Now, substitute and compute**:
\[
\chi^2 = \frac{24 \cdot 64}{289}
\]
\[
\chi^2 = \frac{1536}{289}
\]
\[
\chi^2 \approx 5.31
\]

### Test Statistic:
The test statistic is approximately \(\chi^2 \approx 5.31\).

### Compare Test Statistic with Critical Value:
We previously found the critical value for \(df = 24\) at the \(\alpha = 0.10\) level of significance to be approximately \(33.196\).

Since the test statistic \(5.31\) is less than the critical value \(33.196\), we fail to reject the null hypothesis. Therefore, there is insufficient evidence to support the claim that the population standard deviation is greater than 17 at the 0.10 level of significance.