the mean of adult men is 172 pounds with a standard deviation of 29 pounds. ask 20 men thier weight and use excel to calculate the mean. test the hypothesis that the mean is not 172 pounds at the .10 significance level. and how many men should be sampled if you want to be 98% confident that the mean will be within an error of 3lbs.
We have no access to your sample data.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z score.
Remember that:
Ho: mean = 172
Ha: mean ≠ 172
To test the hypothesis that the mean weight of adult men is not 172 pounds at the 0.10 significance level, we can use the following steps:
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The mean weight of adult men is 172 pounds.
- Alternative hypothesis (H1): The mean weight of adult men is not 172 pounds.
Step 2: Collect data from 20 men and calculate the sample mean using Excel. Let's assume the weights of the 20 men are as follows:
180, 165, 175, 185, 170, 160, 175, 170, 180, 190, 175, 165, 170, 180, 185, 190, 195, 170, 175, 180.
Using Excel to calculate the mean, we add up all the weights and divide by the number of men (20):
Mean = (180 + 165 + 175 + 185 + 170 + 160 + 175 + 170 + 180 + 190 + 175 + 165 + 170 + 180 + 185 + 190 + 195 + 170 + 175 + 180) / 20
Mean = 3,535 / 20
Mean = 176.75 pounds
The sample mean weight is calculated as 176.75 pounds.
Step 3: Calculate the test statistic (t-value) using Excel. Since the population standard deviation is known (29 pounds), we can use a one-sample t-test formula:
t = (Sample Mean - Population Mean) / (Population Standard Deviation / √Sample Size)
t = (176.75 - 172) / (29 / √20)
t = 4.75 / (29 / 4.472136)
t ≈ 4.75 / 6.47343
t ≈ 0.732
The calculated t-value is approximately 0.732.
Step 4: Determine the critical t-value. Since the significance level is 0.10, we need to find the critical t-value for a two-tailed test with 20-1 = 19 degrees of freedom. Using a t-table or Excel's T.INV function, the critical t-value at a 0.10 significance level is approximately ±1.729.
Step 5: Compare the calculated t-value with the critical t-value. If the calculated t-value falls within the range of the critical t-value, we fail to reject the null hypothesis. Otherwise, we reject the null hypothesis.
In this case, 0.732 falls within the range of ±1.729. Therefore, we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean weight of adult men is significantly different from 172 pounds at the 0.10 significance level.
Now, let's move on to the second question regarding the sample size needed to be 98% confident that the mean will be within an error of 3 pounds.
To calculate the required sample size, we can use the formula:
n = (Z * σ / E)^2
Where:
n = required sample size
Z = Z-value corresponding to the desired confidence level (98% confidence level corresponds to a Z-value of approximately 2.326 from the standard normal distribution table)
σ = population standard deviation
E = margin of error (3 pounds)
Plugging in the values:
n = (2.326 * 29 / 3)^2
n ≈ (67.9874)^2
n ≈ 4625.85
The required sample size should be approximately 4626 men to be 98% confident that the mean weight will be within an error of 3 pounds. However, since it is not possible to have a fraction of a person, round up to the nearest whole number, so the final sample size needed would be 4626 men.
To test the hypothesis that the mean weight of adult men is not 172 pounds, you can use a one-sample t-test. Here are the steps to perform this test in Excel:
1. Enter the weights of the 20 men in a column in Excel.
2. Calculate the mean and standard deviation of the sample using Excel functions. Let's assume the weights are in cells A1 to A20, you can use the following formulas:
- Mean: =AVERAGE(A1:A20)
- Standard Deviation: =STDEV(A1:A20)
3. Calculate the t-statistic, which measures the difference between the sample mean and the hypothesized mean (172 pounds). Use the formula:
- =ABS(Mean - 172) / (Standard Deviation / SQRT(20)), where Mean is the calculated mean and Standard Deviation is the calculated standard deviation.
4. Determine the critical t-value for a significance level of 0.10 (90% confidence level). Use the T.INV function in Excel:
- =T.INV(0.10, 19) - this returns the critical t-value for a two-tailed test with 19 degrees of freedom (20 - 1).
5. Compare the calculated t-statistic from step 3 with the critical t-value from step 4. If the calculated t-statistic is greater than the critical t-value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.
Now, to determine the sample size needed to be 98% confident that the mean weight will be within an error of 3 pounds, you can use the following formula:
n = (Z * σ / E)^2
where:
n = required sample size
Z = Z-score corresponding to the desired confidence level (in this case, 98% confident, so Z ≈ 2.33 based on a standard normal distribution table)
σ = standard deviation of the population (given as 29 pounds)
E = maximum acceptable error (3 pounds)
Plugging in the values, you can calculate the required sample size:
n = (2.33 * 29 / 3)^2 ≈ 75.19
Therefore, you would need to sample at least 76 men to be 98% confident that the mean weight will be within an error of 3 pounds. Since you cannot have a fraction of a person, round up to the nearest whole number, resulting in 76 men.