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In: Statistics and Probability

John measured the total playing time of 35 randomly chosen CDs from his very large collection...

John measured the total playing time of 35 randomly chosen CDs from his very large collection and found a mean of 79.7 minutes and a standard deviation of 9.8 minutes. Give the 93% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places.

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Expert Solution

Solution :

Given that,

= 79.7

= 9.8

n = 35

At 93% confidence level the z is ,

  = 1 - 93% = 1 - 0.93 = 0.07

/ 2 = 0.07 / 2 = 0.035

Z/2 = Z0.035 = 1.81

Margin of error = E = Z/2* ( /n)

= 1.81 * (9.8 / 35)

= 3.00

At 93% confidence interval estimate of the population mean is,

- E < < + E

79.7 - 3.00 < < 79.7 + 3.00

76.70 < < 82.70

(76.70 , 82.70)

orchestra answered 2 years ago
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