Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.ÂÂ
23) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that CJ 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.ÂÂ
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.ÂÂ
Do one of the following, as appropriate: (a) Find the critical value Za/2- (b) find the critical value ta/2- (c) state thatÂÂ
neither the normal nor the t distribution applies.ÂÂ
24) 99%; n = 17; CJ is unknown; population appears to be normally distributed.ÂÂ
A) Za./2 = 2.583 B) Za/2 = 2.567 C) ta/2 = 2.898 D) ta./2 = 2.921ÂÂ
Use the given degree of confidence and sample data to construct a confidence interval for the population mean 1-1· AssumeÂÂ
that the population has a normal distribution.ÂÂ
25) The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:ÂÂ
7.0 10.8 9.5 8.0 11.5ÂÂ
7.5 6.4 11.3 10.2 12.6ÂÂ
Determine a 95% confidence interval for the mean time for all players (include a means plot for the data)ÂÂ
A) 7.93 min<>< 11.03=””>
C) 10.93 min< 1-1=””>< 8.03=””>
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B) 8.03 min< 1-1=””>< 10.93=””>
D) 11.03 min< 1-1=””>< 7.93=””>
The Correct Answer and Explanation is :
Let’s address each part of the problem step by step:
1. Null Hypothesis (H₀) and Alternative Hypothesis (H₁):
- Null Hypothesis (H₀): The population mean weight of all employees is 200 lb.
- H₀: μ = 200
- Alternative Hypothesis (H₁): The population mean weight of all employees is less than 200 lb.
- H₁: μ < 200
2. Test Statistic:
Since the population standard deviation (σ) is known, we use the z-test for the population mean. The test statistic (z) is calculated as:
[ z = \frac{\overline{x} – \mu}{\frac{\sigma}{\sqrt{n}}} ]
Where:
- (\overline{x}) = sample mean = 183.9 lb
- μ = hypothesized population mean = 200 lb
- σ = population standard deviation = 121.2 lb
- n = sample size = 54
Calculating the standard error:
[ \frac{\sigma}{\sqrt{n}} = \frac{121.2}{\sqrt{54}} \approx 16.49 \, \text{lb} ]
Now, calculating the z-score:
[ z = \frac{183.9 – 200}{16.49} \approx -0.976 ]
3. P-value:
The p-value represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. For a z-score of -0.976, we look up the corresponding cumulative probability in the standard normal distribution table.
Using standard normal distribution tables or a calculator, the cumulative probability for z = -0.976 is approximately 0.164.
Since this is a one-tailed test (μ < 200), the p-value is 0.164.
4. Conclusion about the Null Hypothesis:
The significance level (α) is 0.10. Since the p-value (0.164) is greater than α (0.10), we fail to reject the null hypothesis.
5. Final Conclusion:
There is insufficient evidence at the 0.10 significance level to support the claim that the population mean weight of all employees is less than 200 lb.
Regarding the Multiple Choice Question:
The question asks for the critical value for a 99% confidence level with a sample size of 17, assuming the population appears to be normally distributed and the population standard deviation is unknown.
When the population standard deviation is unknown and the sample size is small (n < 30), we use the t-distribution. The degrees of freedom (df) are calculated as:
[ \text{df} = n – 1 = 17 – 1 = 16 ]
For a 99% confidence level, the critical value corresponds to the 0.005 quantile of the t-distribution with 16 degrees of freedom. Using a t-distribution table or calculator, the critical value is approximately 2.921.
Therefore, the correct answer is:
D) t₀.005,16 = 2.921
Confidence Interval Calculation:
To construct a 99% confidence interval for the population mean, we use the formula:
[ \overline{x} \pm t_{\alpha/2, \, \text{df}} \times \frac{s}{\sqrt{n}} ]
Where:
- (\overline{x}) = sample mean
- (t_{\alpha/2, \, \text{df}}) = critical value from the t-distribution
- (s) = sample standard deviation
- (n) = sample size
Since the specific sample data (mean and standard deviation) are not provided in the question, we cannot compute the exact confidence interval. However, the critical value for a 99% confidence level with 16 degrees of freedom is 2.921, as determined above.
Regarding the Football Drill Times:
The question provides the following drill times (in minutes): 7.0, 10.8, 9.5, 8.0, 11.5, 7.5, 6.4, 11.3, 10.2, 12.6.
To determine a 95% confidence interval for the mean time for all players, we follow these steps:
- Calculate the Sample Mean ((\overline{x})): [ \overline{x} = \frac{7.0 + 10.8 + 9.5 + 8.0 + 11.5 + 7.5 + 6.4 + 11.3 + 10.2 + 12.6}{10} = 9.33 \, \text{minutes} ]
- Calculate the Sample Standard Deviation (s): First, compute the squared deviations from the mean: [
(7.0 – 9.33)^2 = 5.4289, \quad (10.8 – 9.33)^2 = 2.1769, \quad (9.5 – 9.33)^2 = 0.0289, \quad (8.0 – 9.33)^2 = 1.7689, \quad (11.5 – 9.33)^2 = 4.7089
]
[
(7.5 – 9.33)^2 = 3.4289, \quad (6.4 – 9.33)^2 = 8.5289, \quad (11.3 – 9.33)^2 = 3.9289, \quad (10.2 – 9.33)^2 = 0.7569, \quad (12.6