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.

135 131 191
168 214 139
170 197
136 178
179 154
183 158
149 200
152 211
271 186

For the first part of this problem, you can try a z-test. You will need to calculate the sample mean before using the formula below:

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

Population mean = 172
Standard deviation = 29
Sample size = 20

Once you have the z-test statistic, check a z-table for a two-tailed test at .10 level of significance. If the test statistic exceeds the critical value in either tail, reject the null. If the test statistic does not exceed the critical value in either tail, do not reject the null.

For the second part, use a formula to find sample size:

Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.33 using a z-table to represent the 98% confidence interval, sd = 29, E = 3, ^2 means squared, and * means to multiply.

Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.

Hope this helps.

To test the hypothesis that the mean is not 172 pounds at a significance level of 0.10, you can perform a one-sample t-test using Excel. Here are the steps to calculate the mean and conduct the hypothesis test:

1. Enter the 20 men's weights into a column in an Excel spreadsheet.
2. In an empty cell, use the function "=AVERAGE()" and select the range of weights to calculate the sample mean. This will give you the mean weight of the 20 men.
3. Calculate the standard deviation of the sample using the "=STDEV()" function and the range of weights.
4. Compute the t-score using the formula: t = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size)), where the sample mean is the mean weight calculated in step 2, the hypothesized mean is 172 pounds, the standard deviation is the value given in the question (29 pounds), and the sample size is 20.
5. Use the "=T.DIST()" function to calculate the p-value for the t-score obtained. The formula should be "=T.DIST(abs(t-score), degrees of freedom, 1)", where abs(t-score) is the absolute value of the t-score obtained in step 4, and degrees of freedom is the sample size minus one (19 in this case).
6. Finally, compare the p-value obtained with the significance level (0.10) to determine whether to reject or fail to reject the null hypothesis. If the p-value is less than 0.10, reject the null hypothesis and conclude that the mean weight is not 172 pounds. If the p-value is greater than or equal to 0.10, fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the mean weight is different from 172 pounds.

Regarding the second part of your question, to determine how many men should be sampled if you want to be 98% confident that the mean will be within an error of 3 pounds, you can use the formula for the sample size required for estimating a population mean with a given confidence interval.

The formula is given by:
n = (Z * σ / E)^2

Where:
n = required sample size
Z = z-value corresponding to the desired confidence level (98% confidence corresponds to approximately 2.33 z-value)
σ = standard deviation of the population (given as 29 pounds)
E = maximum error (given as 3 pounds)

Plugging in the values, you get:
n = (2.33 * 29 / 3)^2

Calculating this equation will give you the minimum sample size required.