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:

The t-score we use for the confidence interval is two-tailed , i.e. the  t-score should be used to find a ($1-\alpha$) is given by :-

$t_{n-1, \alpha/2}$, where n is the sample size.

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

Significance level : $\alpha: 1-0.80=0.20$

Sample size : n= 47

Then, degree of freedom : $n-1=46$

Now by using standard normal t-distribution table,

$t_{n-1, \alpha/2}=t_{46, 0.10}=1.300228$

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