Confidence interval for when variance is known confidence interval for when vari

Question

Confidence interval for when variance is known confidence interval for when variance of is unknown 1. Assume that the distribution of the incomes is approximately normal. We take a sample of 36 students from Penn State University and record their family incomes. Suppose the sample average income is $80,000 and sample standard deviation is $10,000. Construct a 95% confidence interval for the true average income a) Construct a 95% confidence interval for the true average income [3 points] If the confidence level increase to 99%, the confidence interval for will be wider or narrower? Please demonstrate the 99% confidence interval to support your conclusion. [2 points] )

Explanation / Answer

a)

M = 80000
t = 1.96
sM = (100002/36) = 1666.67

= M ± Z(sM)
= 80000 ± 1.96*1666.67
= 80000 ± 3266.61

Confidence Interval = [76733.39, 83266.61]

b)

The confidence interval will be wider.

M = 80000
t = 2.58
sM = (100002/36) = 1666.67

= M ± Z(sM)
= 80000 ± 2.58*1666.67
= 80000 ± 4293.05

Confidence Interval = [75706.95, 84293.05]

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