To address the given question, let's take it step by step by examining the provided data and understanding the statistical concepts involved.
### Given Data:
- Sample mean ([tex]$\bar{x}$[/tex]): 183.9 lb
- Population mean ([tex]$\mu$[/tex]): 200 lb
- Standard deviation ([tex]$\sigma$[/tex]): 121.2 lb
- Sample size (n): 54
- Significance level ([tex]$\alpha$[/tex]): 0.10
- Test statistic (z): -0.98
- P-value: 0.1635
### Hypotheses:
- Null hypothesis ([tex]$H_0$[/tex]): The population mean is 200 lb ([tex]$\mu = 200$[/tex]).
- Alternative hypothesis ([tex]$H_1$[/tex]): The population mean is less than 200 lb ([tex]$\mu < 200$[/tex]).
### Interpretation of the P-value:
The P-value is a measure of the probability that the observed test statistic would occur under the null hypothesis. If the P-value is less than the significance level, we reject the null hypothesis. In this case:
- Significance level ([tex]$\alpha$[/tex]): 0.10
- P-value: 0.1635
Since the P-value (0.1635) is greater than the significance level (0.10), we fail to reject the null hypothesis. This means that there is not enough statistical evidence to support the claim that the population mean weight of employees is less than 200 lb.
### Conclusion:
Based on our comparison of the P-value to the significance level:
- We fail to reject the null hypothesis ([tex]$H_0$[/tex]).
- Therefore, there is not sufficient evidence to support the claim that the population mean weight of employees is less than 200 lb.
### Final Conclusion:
Given the data and the results of our hypothesis test, the final conclusion is that we fail to reject the null hypothesis. Hence, there is not sufficient evidence to support the claim that the mean weight of the employees is less than 200 pounds.
The correct answer is:
P-value: 0.1635. Fail to reject [tex]$H_0$[/tex]. There is not sufficient evidence to support the claim that the mean is less than 200 pounds.
