There are 300 students enrolled in Business Statistics. Historically, exam scores are normally distributed with a...

There are 300 students enrolled in Business Statistics. Historically, exam scores are normally distributed with a standard deviation of 26.7. Your instructor randomly selected a sample of 40 examinations and finds a mean of 64.2. Determine a 90% confidence interval for the mean score for all students taking the course. 90% CI = to

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Expert Solution

Solution :

Given that,

= 64.2

= 26.7

n = 40

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( / $\sqrt$ n)

= 1.645 * ( 26.7/ $\sqrt$ 40)

= 6.9446

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

- E < < + E

64.2 - 6.9446 < < 64.2 + 6.9446

57.2554< < 71.1446

(57.2554 to 71.1446 )

orchestra answered 1 year ago

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