Question

Assume that a sample is used to estimate a population mean μμ. Find the 99.9% confidence interval for a sample of size 65 with a mean of 37.9 and a standard deviation of 15.4. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
99.9% C.I. =_____________.

Answer 1

Answer:

C.I 99%(μ)= [10.3; 20.5]

Step-by-step explanation:

The confidence interval formula is:  

I (1-alpha) (μ)= mean+- [(t(n-1))* S/sqrt(n)]  

alpha= is the proportion of the distribution tails that are outside the confidence interval. In this case, 1% because 100-99%  

n= number of observations = 65

t(n-1)= is the critical value of the t distribution with 65-1 degrees of freedom for an area of alpha/2 (0.5%). In this case is 2.6603

We use the t-student distribution because the population standard deviation is unknown.  

S= sample standard deviation. In this case 15.4

mean= 37.9

Then, the confidence interval (99%):  

C.I 99%(μ)= 15.4+- [2.6603*(15.4/sqrt(65)]  

C.I 99%(μ)= 15.4+- [5.0815]  

C.I 99%(μ)= [15.4-5.0815; 15.4+5.0815]  

C.I 99%(μ)= [10.3185; 20.4815]  

C.I 99%(μ)= [10.3; 20.5]