a sociologist develops a test designed to measure attitudes about disabled people and gives the test to 16 randomly selected subjects. their mean score is 71.2 with a standard deviation of 10.5. construct the 99 percent confidence interval for the mean score of all subjects.

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rafikiedu08 rafikiedu08 · Answer 1 · 2022-12-06T12:26:28+01:00

The 99 percent confidence interval for the mean score of all subjects lies greater than 63.464 and less than 78.936.

Given:

No. of samples (n)= 16

df = n-1 = 16 -1 = 15

The mean score is 71.2

The standard deviation = 10.5.

So, from the t-distribution of critical values table through the above-mentioned details. We get the values:

Alpha / 2 = 0.005

z = 99/100 + 0.005 = .99 + .005 = .995

Sample mean = xbar = 76.2

t = 2.947

The formula for standard error = E = value of t (at z, and df) / S.D./sqrt(n)

i.e. E = 2.947 x 10.5 / sqrt 16 = 7.736

I = 71.2 +/- E = 71.2 +/- 7.736

99 percent confidence interval - 71.2 +/- 7.736

Adding both sides 7.736 for range. We get,

63.464 < u < 78.936

To learn more about t-distribution visit: https://brainly.com/question/14860724

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