Thanks to an initiative to recruit top students, an administrator at a college claims that this year's entering class must have a greater mean IQ score than that of entering classes from previous years. The administrator tests a random sample of 15 of this year's entering students and finds that their mean IQ score is 111, with a standard deviation of 12. The college records indicate that the mean IQ score for entering students from previous years is 110.

Is there enough evidence to conclude, at the 0.05 level of significance, that the population mean IQ score, μ, of this year's class is greater than that of previous years? To answer, assume that the IQ scores of this year's entering class are approximately normally distributed.

Perform a one-tailed test. Then complete the parts below.

Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas)

(A) State the null hypothesis H and the alternative hypothesis II

(b) Determine the type of test statistic to use.

(Choose one) ▼

[C] Find the value of the test statistic (Round to three or more decimal places.)

(D) Find the p-value. (Round to three or more decimal places.)

(e) Can we conclude that the mean 10 score of this year's class is greater than that of previous years?

Let's walk through the process of hypothesis testing step by step.

### (A) State the Null Hypothesis (H₀) and the Alternative Hypothesis (H₁):

- **Null Hypothesis (H₀):** The mean IQ score of this year's entering class is equal to or less than that of previous years.
  \[
  H_0: \mu = 110
  \]
- **Alternative Hypothesis (H₁):** The mean IQ score of this year's entering class is greater than that of previous years.
  \[
  H_1: \mu > 110
  \]

### (B) Determine the Type of Test Statistic to Use:

Given that the sample size is small (n = 15) and the population standard deviation is unknown, we should use the **t-test** for the test statistic.

### (C) Find the Value of the Test Statistic:

The test statistic for a one-sample t-test is calculated using the following formula:
\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]
Where:
- \(\bar{x}\) = sample mean = 111
- \(\mu_0\) = population mean from previous years = 110
- \(s\) = sample standard deviation = 12
- \(n\) = sample size = 15

Substituting the values:
\[
t = \frac{111 - 110}{\frac{12}{\sqrt{15}}} = \frac{1}{\frac{12}{\sqrt{15}}}
\]

Let's calculate this value.

\[
t = \frac{1}{\frac{12}{\sqrt{15}}} \approx \frac{1}{3.098} \approx 0.323
\]
\[
t \approx 0.323
\]

### (D) Find the p-Value:

To find the p-value, we need to look up the t-statistic in the t-distribution table or use statistical software. Since this is a one-tailed test and our alternative hypothesis is \(H_1: \mu > 110\), the p-value corresponds to the area to the right of the calculated t-statistic.

For \(df = n - 1 = 15 - 1 = 14\):

Let's calculate the p-value using the t-distribution.

\[
p\text{-value} = P(T > t) 
\]

Using the t-distribution, we find:

\[
\text{p-value} \approx 0.374
\]

### (E) Conclusion:

At the 0.05 level of significance, we compare the p-value with \(\alpha = 0.05\):

- If \(p\text{-value} \leq \alpha\), we reject the null hypothesis.
- If \(p\text{-value} > \alpha\), we fail to reject the null hypothesis.

Since \(0.374 > 0.05\), we **fail to reject** the null hypothesis.

**Conclusion:** There is not enough evidence to conclude that the mean IQ score of this year's class is greater than that of previous years.

---

**Do you want details or have any questions?**

Here are some relative questions to consider:

1. What would happen if the p-value was less than the significance level?
2. How does the sample size affect the power of the test?
3. What assumptions are required for the t-test to be valid?
4. How would the result differ if we used a two-tailed test instead of a one-tailed test?
5. Why is the t-test used instead of a z-test in this scenario?

**Tip:** Always ensure that the assumptions of normality and sample size adequacy are met before conducting a t-test, especially when working with small samples.

Explore hypothesis testing to determine if this year's entering class has a greater mean IQ score than previous years. Includes step-by-step calculations using a one-sample t-test with detailed explanations of the null and alternative hypotheses, test statistic calculation, and interpretation of results.

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