Suppose that a random sample of 11 adults has a mean score of 71 on a standardiz

Question

Suppose that a random sample of 11 adults has a mean score of 71 on a standardized personality test, with a standard deviation of 4. (A higher score indicates a more personable participant.) If we assume that 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 least three decimal places. Round your answers to one decimal place. what is the lower and upper limit of confidence interval

Explanation / Answer

confidence intervel = x+/- z*s/squareroot (n)

=71+/- 2.576*4/squareroot 11

=71+/- 2.576 *4/3.316624790

=71+/- 3.106772895

67.9 ,74.1

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