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?
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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.