Question

An engineer who is studying the tensile strength of a steel alloy knows that tensile strength
is approximately normally distributed with σ = 60 psi. A random sample of 12 specimens
has a mean tensile strength of 3450 psi.

a) Compute a two-sided confidence interval on the mean tensile strength at a 95%
confidence level

b) Test the hypothesis that the mean strength is 3500 psi at α= 0.05, and provide a
conclusion statement

Answer 1

a) The 95% confidence interval for population mean

Degrees of freedom = n-1 = 12-1 =11

For 95% confidence (with df = 11 , two tailed critical value of t is

tc = 2.20 ( from t table )

Thus , 95% confidence interval for mean textile strength

=(3411.90 psi, 3488.11 psi)

Note Though population standard deviation is given , we use t distribution as sample size is small, and we use in place of s

b)  The null and alternative hypothesis

3500

3500

Test statistic

= - 2.89

For with df = 11 , two tailed critical value of t is

tc = 2.20 ( from t table )

Since calculated value of test statistic , I t I > 2.20

We reject H0

At 5% level of significance there is sufficient evidence to conclude that mean textile strength is not equal to 3500 psi .