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.
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
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