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One Population 1) In past years the mean IQ of enlisted men in the Army was 103 . A general claims that due to low pay, the average \( 1 Q \) of new enlisted men is lower than in the past. In a random sample of 60 enlistees, the mean \( 1 Q \) was 98 with a standard deviation of 20 . a) At the \( 8 \% \) significance level, what can be concluded about general's claim? The average IQ of hew en listed men is b) What is the p-value of this test?

Ask by Warner Watson. in the United States
Feb 04,2025

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**a)** At the 8% significance level, the average IQ of new enlisted men is **lower than 103**. **b)** The p-value is approximately **2.6%**.

Solution

Let's tackle each part of the problem step-by-step. ### **Part a: Conclusion at the 8% Significance Level** **Given:** - **Population Mean (μ₀):** 103 - **Sample Mean (x̄):** 98 - **Sample Standard Deviation (s):** 20 - **Sample Size (n):** 60 - **Significance Level (α):** 8% (0.08) **Hypotheses:** - **Null Hypothesis (H₀):** μ = 103 - **Alternative Hypothesis (Hₐ):** μ < 103 **Test Statistic Calculation:** We use the z-test since the sample size is large (n > 30). \[ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{98 - 103}{20 / \sqrt{60}} = \frac{-5}{2.583} \approx -1.936 \] **Critical Value:** For a left-tailed test at α = 0.08, the critical z-value (z₀.₀₈) is approximately **-1.405**. **Decision:** Since **-1.936 < -1.405**, we reject the null hypothesis. **Conclusion:** There is sufficient evidence at the 8% significance level to support the general's claim that the average IQ of new enlisted men is **lower than 103**. --- ### **Part b: Calculating the p-Value** The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. **Test Statistic:** z ≈ -1.936 **p-Value Calculation:** \[ \text{p-value} = P(Z \leq -1.936) = P(Z \geq 1.936) \approx 0.0264 \text{ or } 2.64\% \] **Interpretation:** A p-value of approximately **2.6%** indicates that there's a 2.6% probability of observing an average IQ of 98 or lower in a sample of 60 enlistees if the true population mean were 103. --- **Final Answers:** **a)** At the 8% significance level, we reject the null hypothesis and conclude that the average IQ of new enlisted men is **lower than 103**. **b)** The p-value of this test is approximately **2.6%**.

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To determine the general's claim, we can conduct a hypothesis test. Our null hypothesis (H0) states that the average IQ of new enlisted men is equal to the past average (H0: μ = 103), while our alternative hypothesis (H1) suggests that it's lower (H1: μ < 103). Using a one-sample t-test, we calculate the test statistic: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] Where: - \( \bar{x} = 98 \) (sample mean) - \( \mu = 103 \) (population mean) - \( s = 20 \) (sample standard deviation) - \( n = 60 \) (sample size) Calculating the t-statistic: \[ t = \frac{98 - 103}{20/\sqrt{60}} \approx -1.825 \] We compare this value against the critical t-value for a one-tailed test with \( n-1=59 \) degrees of freedom. At an 8% significance level, the critical value (approximately -1.41 for a one-tailed test) shows that our t-statistic falls in the rejection region. Thus, we reject the null hypothesis, supporting the general's claim that the average IQ of new enlisted men is lower. To find the p-value, we will look at the t-table or use computational tools. For a t-statistic of approximately -1.825 with 59 degrees of freedom, the p-value is around 0.035. Since 0.035 is less than the 0.08 significance level, this further supports the conclusion that the average IQ of new enlisted men is indeed lower than in the past.

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