Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. Choose the correct option based on the problem information.

Question

A large software company gives job applicants a test of programming ability and the mean for that test has been 160 in the past. Twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160. Use the P-value method of testing hypotheses. Choose the correct P-value and conclusion based on the null hypothesis.

A) P-value = 0.05. Reject H_0.
B) 0.01 < P-value < 0.025. Reject H_0.
C) P-value < 0.005. Reject H_0.
D) P-value < 0.005. Fail to reject H_0.

Answer 1

To test the claim that the sample comes from a population with a mean score greater than 160, we'll go through the following steps:

1. Identify Null and Alternative Hypotheses:

   • Null Hypothesis (\( H_0 \)): \( \mu \leq 160 \)
   • Alternative Hypothesis (\( H_1 \)): \( \mu > 160 \)

2. Test Statistic: We will use the formula for the test statistic for the sample mean from a normally distributed population: \[ z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \] where:

   • \( \bar{x} = 183 \) (sample mean),
   • \( \mu_0 = 160 \) (hypothesized population mean),
   • \( s = 12 \) (sample standard deviation),
   • \( n = 25 \) (sample size).

   Plugging in the values: \[ z = \frac{183 - 160}{\frac{12}{\sqrt{25}}} = \frac{23}{\frac{12}{5}} = \frac{23}{2.4} \approx 9.5833 \]

3. P-value: Using a standard normal distribution (z-table or calculator), we can find the P-value associated with a test statistic of \( z \approx 9.5833 \).

   A z-score this high (greater than 3.49) will have a very small P-value, certainly much less than 0.005. Therefore:

   • \( P \text{-value} < 0.005 \)

4. Conclusion: We compare the P-value to the level of significance (\( \alpha = 0.05 \)):

   • Since \( P \text{-value} < 0.005 \) which is less than \( 0.05 \), we reject the null hypothesis \( H_0 \).

5. Final Statement: Given that we have rejected the null hypothesis, we conclude that there is sufficient evidence to support the claim that the mean score of the population from which this sample is drawn is greater than 160.

Based on the above analysis, the correct option is: C) P-value < 0.005. Reject H_0.