Question

Suppose that the amount spent per person per month on health insurance in the U.S. is normally distributed with an unknown mean and standard deviation. The monthly health insurance costs of 17 randomly sampled individuals in the U.S. are used to estimate the mean of the population. What t-score should be used to find the 90% confidence interval for the population mean?

Answer 1

Answer: 1.745884

Explanation:

Given : Level of significance :
`1-\alpha:0.90`

Then , significance level :
`\alpha: 1-0.90=0.10`

Sample size :
`n=17`

Then , the degree of freedom for t-distribution:
`df=n-1=16`

The t-score for a level of confidence
`(1-\alpha)` is given by the formula below :-

`t_((df,\alpha/2))`, where df is the degree of freedom and
`\alpha` is the significance level.

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

`\text{ t-score}=t_((df,\alpha/2))=t_(16,0.05)=1.745884`

Thus, the t-score should be used to find the 90% confidence interval for the population mean =1.745884