Question

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?

Answer 1

Answer: 1.300228

Step-by-step explanation:

We know that the t-score for a level of confidence $(1-\alpha)$ is given by :_

$t_{(df,\alpha/2)}$, whre df is the degree of freedom and $\alpha$ is the significance level.

Given : Level of significance : $1-\alpha:0.80$

Then , significance level : $\alpha: 1-0.80=0.20$

Since , sample size : $n=47$

Degree of freedom for t-distribution: $df=n-1=47-1=46$

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

$t_{(df,\alpha/2)}=t_{46,0.10}=1.300228$

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