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Identify the P-value, conclusion regarding the null hypothesis, and final conclusion that addresses the original claim.

The health of employees is monitored by periodically recording their weight. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that σ is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. H_0: μ = 200; H_1: μ < 200; Test statistic: z = −0.98.

A) P-value: 0.8365. Reject H_0. There is sufficient evidence to support the claim that the mean is less than 200 pounds.
B) P-value: 0.1635. Fail to reject H_0. There is not sufficient evidence to support the claim that the mean is less than 200 pounds.
C) P-value: 0.1635. Reject H_0. There is not sufficient evidence to support the claim that the mean is less than 200 pounds.
D) P-value: 0.8365. Fail to reject H_0. There is sufficient evidence to support the claim that the mean is less than 200 pounds.

1 answer

To reach a conclusion, we first need to interpret the P-value based on the test statistic given, and then we will apply it to the hypotheses stated.

Given:

  • Null Hypothesis \( H_0: \mu = 200 \)
  • Alternative Hypothesis \( H_1: \mu < 200 \)
  • Test Statistic \( z = -0.98 \)
  • Significance Level \( \alpha = 0.10 \)

A z-score of -0.98 indicates how many standard deviations the sample mean is below the population mean of 200.

Next, we look up the P-value associated with this z-score. The P-value represents the probability of observing a test statistic as extreme as (or more extreme than) the sample statistic, assuming the null hypothesis is true. For a left-tailed test (as indicated by \( H_1 \)), the P-value corresponds to the area to the left of z = -0.98 in the standard normal distribution.

Using a standard normal distribution table or calculator, the P-value for \( z = -0.98 \) is approximately 0.1635.

Now, we compare this P-value to our significance level, \( \alpha = 0.10 \):

  • Since \( P \text{-value} (0.1635) > \alpha (0.10) \), we fail to reject the null hypothesis \( H_0 \).

Final conclusion: There is not sufficient evidence to support the claim that the mean weight of the employees is less than 200 lb.

Thus, the correct choice is: B) P-value: 0.1635. Fail to reject H_0. There is not sufficient evidence to support the claim that the mean is less than 200 pounds.