In a simple random sample of 100 households, the sample mean number of personal computers was...

In a simple random sample of 100 households, the sample mean number of personal computers was 1.32. Assume the population standard deviation is 0.41. Construct a 95% confidence interval for the mean number of personal computers.

(a) (1.24, 1.40)

(b) (1.25, 1.39)

(d) (0.19, 0.63)

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 1.32

Population standard deviation = = 0.41

Sample size = n = 100

At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

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

= 1.96 * ( 0.41 / $\sqrt$ 100 )

= 0.08

At 95% confidence interval estimate of the population mean is,

- E < < + E

1.32 - 0.08 < < 1.32 + 0.08

(1.24 < < 1.40)

(1.24, 1.40)

correct option is = a)

milcah answered 9 months ago

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