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A random sample of 20 houses selected from a city showed that the mean size of these houses is 1880 square feet with a standard deviation of 320 square feet. Assume that the sizes of all houses in this city have an approximate normal distribution. The upper bound the 90% confidence interval for the mean size of all houses in this city is: A. 2110 B. 1941 C. 1974 D. 1968 E. 1894

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Answer:

Option C

Step-by-step explanation:

Given that for a  random sample of 20 houses selected from a city showed that the mean size of these houses is 1880 square feet with a standard deviation of 320 square feet.

X, the sizes of houses is Normal

Hence sample mean will be normal with

Mean = 1880

and std dev = [tex]\frac{320}{\sqrt{20} } \\=71.5542[/tex]

For 90% confidence interval critical value for t with degree of freedom 19 is

1.328

Confidence interval upper bound

= mean + margin of error

= [tex]1880+1.328*71.55\\=1975[/tex]

Option C is right

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