2. Determine the 99.5% CI for the population mean and explain why the Z or T sta

Question

2. Determine the 99.5% CI for the population mean and explain why the Z or T statistic applies.

If the question can’t be answered with the given information, explain why that is so. [Hint:

Calculator’s CI facilities]

2.1 Random sample size 25, mean 50, and standard deviation 16; unknown population

distribution

2.2 Same as 2.1 but with known normal population and known population standard deviation

2.3 Same as 2.1 but sample size 250 with unknown standard deviation and population

believed to be non-symmetric with a considerable skew, extreme outliers, and possible multiple

modes.

Explanation / Answer

General Rule:

You must use the t-distribution table when working problems when the population standard deviation () is not known and the sample size is small (n<30). General Correct Rule: If is not known, then using t-distribution is correct. If is known, then using the normaldistribution is correct.

2.1

df = 24

alpha = 0.005

t-value(alpha/2) = 3.0905

CI = ( 50 - 3.0905 , 50+3.0905) = (46.9095 , 53.9095) Answer

2.2

z for 99.5 % = 2.807

Margin of error = (z*SD)/sqrt(25)

= ( 2.807 * 16) / sqrt(25)

= 8.9824

CI = ( 50 - 8.9824 , 50+8.9824) = (41.0176 , 58.0176) Answer

2.3

df = 249

alpha = 0.005

t-value(alpha/2) = 2.83(approx)

CI = ( 50 - 2.83 , 50+2.83) = (47.17 , 52.83) Answer

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