Question

The average of a random sample of 10 scores on a college placement exam is x ¯= 75, and the sample standard deviation is s = 8.4. Assuming that the collection of all scores on the exam is approximately normally distributed, find a 90 percent confidence interval for the mean score.

Answer #1

Solution :

Given that,

t /2,df = 1.833

Margin of error = E = t/2,df * (s /n)

= 1.833 * (8.4 / 10)

Margin of error = E = 4.9

The 90% confidence interval estimate of the population mean is,

- E < < + E

75 - 4.9 < < 75 + 4.9

70.1 < < 79.9

**(70.1 , 79.9)**

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