He number of square feet per house are normally distributed with an unknown population mean and standard deviation. if a random sample of 47 houses is taken to estimate the mean house size, what t-score should be used to find a 80% confidence interval estimate for the population mean?

1.300228

Step-by-step explanation:

We know that the t-score for a level of confidence is given by :_

, whre df is the degree of freedom and is the significance level.

Given : Level of significance :

Then , significance level :

Since , sample size :

Degree of freedom for t-distribution:

With the help of the normal t-distribution table, we have

Hence, the t-score should be used to find a 80% confidence interval estimate for the population mean = 1.300228

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