Posted by Jennifer on Thursday, April 21, 2011 at 2:57pm.
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.
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