Q:

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?

Accepted Solution

A:
Answer: 1.300228Step-by-step explanation:We know that the t-score for a level of confidence [tex](1-\alpha)[/tex] is given by :_[tex]t_{(df,\alpha/2)}[/tex], whre df is the degree of freedom and [tex]\alpha[/tex] is the significance level.Given : Level of significance : [tex]1-\alpha:0.80[/tex]Then , significance level : [tex]\alpha: 1-0.80=0.20[/tex]Since , sample size : [tex]n=47[/tex]Degree of freedom for t-distribution: [tex]df=n-1=47-1=46[/tex]With the help of the normal t-distribution table, we have [tex]t_{(df,\alpha/2)}=t_{46,0.10}=1.300228[/tex]Hence, the t-score should be used to find a 80% confidence interval estimate for the population mean = 1.300228