Question

The following information was obtained from a sample of 37 homes on the market. The square footage of each home was recorded.

Mean square footage: x =1796.4

Sample Standard deviation: s =120.3

A statistics student used the sample information to estimate with 95% confidence by creating a Z interval. Was this the appropriate procedure to use? Why or why not? Explain in words.

Answer 1

No, this was not an appropriate procedure to use as the population standard deviation is not given so we should use a one sample t interval instead of z interval.

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x̅ = 1796.4, s = 120.3, n = 37

95% Confidence interval :

At α = 0.05 and df = n-1 = 36, two tailed critical value, t-crit = T.INV.2T(0.05, 36) = 2.028

Lower Bound = x̅ - t-crit*s/√n = 1796.4 - 2.028 * 120.3/√37 = 1756.29

Upper Bound = x̅ + t-crit*s/√n = 1796.4 + 2.028 * 120.3/√37 = 1836.51

1756.29 < µ < 1836.51