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(16.19) A class survey in a large class for first-year college students asked, "About how many...

(16.19) A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 427 students was x¯¯¯ = 17 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 8 hours in the population of all first-year students at this university. What is the 99% confidence interval (±0.001) for the population mean? Confidence interval is from to hours.

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Expert Solution

solution

Given that,

= 17

= 8

n = 427

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576* (8 / 427)

= 0.997

At 99% confidence interval estimate of the population mean is,

- E < < + E

17-0.997< < 17+0.997

16.003< < 17.997

(16.003,17.997 )

. What is the 99% confidence interval (±0.001) for the population mean

(16.003,17.997 )

milcah answered 8 months ago
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