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.

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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.