Intelligence quotient (IQ) test scores are believed to have a mean 100 of and a population standard deviation of 15. In a random sample of students in a high school, the mean IQ test score is 105 . Researchers claim that the mean IQ test scores at this high school is statistically higher than 100 . What is the Z score? should the null hypothesis be rejected?

To address this problem, we can use **hypothesis testing** and calculate the **Z-score** to determine if the difference in the mean IQ scores is statistically significant. Here's the step-by-step process:

### Given Data:
- Population mean (\(\mu\)): 100
- Population standard deviation (\(\sigma\)): 15
- Sample mean (\(\bar{x}\)): 105
- Sample size (\(n\)): Not provided, but I'll assume it unless otherwise stated.

### Step 1: Set Up Hypotheses
- **Null Hypothesis** (\(H_0\)): The mean IQ score of the students is equal to 100, i.e., \(H_0: \mu = 100\).
- **Alternative Hypothesis** (\(H_a\)): The mean IQ score of the students is higher than 100, i.e., \(H_a: \mu > 100\) (right-tailed test).

### Step 2: Calculate the Z-score

The formula for the Z-score in hypothesis testing is:

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

Where:
- \(\bar{x}\) = Sample mean (105)
- \(\mu\) = Population mean (100)
- \(\sigma\) = Population standard deviation (15)
- \(n\) = Sample size

Since the sample size \(n\) is not provided, we can only solve for the Z-score if we know this value. For now, I’ll proceed as though \(n\) is provided. Let's assume a sample size of 30 for illustration. You can substitute the correct value if necessary.

### Step 3: Z-score Calculation (assuming \(n = 30\))

Using \(n = 30\):

\[
Z = \frac{105 - 100}{\frac{15}{\sqrt{30}}}
\]

Let's calculate that now.

The calculated Z-score is approximately **1.83**.

### Step 4: Determine Whether to Reject the Null Hypothesis

To determine whether to reject the null hypothesis, we compare the Z-score to the critical value for a right-tailed test at a chosen significance level (\(\alpha\)).

- At a common significance level of \( \alpha = 0.05 \), the critical value for a one-tailed test is **1.645**.
- Since the Z-score (1.83) is **greater than 1.645**, we **reject the null hypothesis**.

### Conclusion:
Based on the Z-score of 1.83 and the comparison with the critical value, we reject the null hypothesis. This suggests that the mean IQ score of the students at this high school is **statistically significantly higher** than 100.

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Would you like further details or have any questions about the process? Here are 5 related questions you might explore:

1. How would the Z-score change if the sample size were larger?
2. What is the critical value at different significance levels (e.g., \(\alpha = 0.01\))?
3. How does the population standard deviation affect the Z-score?
4. What happens if we conduct a two-tailed test instead of a one-tailed test?
5. How does a Type I error relate to this hypothesis test?

**Tip:** Larger sample sizes generally reduce the variability of the sample mean, leading to a higher Z-score for the same sample mean.

Intelligence quotient (IQ) test scores are believed to have a mean of 100 and a population standard deviation of 15. In a random sample of students in a high school, the mean IQ test score is 105. Researchers claim that the mean IQ test scores at this high school are statistically higher than 100. What is the Z score? Should the null hypothesis be rejected?

Learn how to test the hypothesis that high school students' IQ scores are statistically higher than the population mean of 100. This problem involves calculating the Z-score using hypothesis testing, normal distribution, and the Central Limit Theorem. Suitable for students in Grades 11-12 and college-level statistics.

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