Suppose that a random sample of 10 adults has a mean score of 76 on a standardiz

Question

Suppose that a random sample of 10 adults has a mean score of 76 on a standardized personality test, with a standard deviation of 6. (A higher score indicates a more personable participant) If we assume scores on this test are normally distributed, find a 99% confidence interval for the mean score of all takers of this test. Then complete the table below. Carry your intermediate computations to at toast three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?

Explanation / Answer

We have been given all params of normal distribution:

n = 10

Mean = 76

Stdev = 6

Z = 99%

The lower limit of CI is 76 - 2.56*6/sqrt(10) = 71.143

The upper limit of CI is 76 + 2.56*6/sqrt(10) = 80.857

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