Question

Pr。Ыет 12. An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed. A random sample of 12 specimens has a mean tensile strength of 3250 psi and a sample standard deviation of 8-60 psi. a) Test the hypothesis that mean strength is 3500 psi. Use α-001. b) What is the smallest level of significance at which you coulji be willing to reject the null hypothesis? c) Build a 95% confidence interval for the mean tensile strength. Does it lead to any conclusion about part a)?

Answer 1

Ans:

a)

Test statistic:

t=(3250-3500)/(60/SQRT(12))

t=-14.434

df=12-1=11

p-value=tdist(14.434,11,2)=0.0000

As,p-value is less than 0.01,we reject the null hypothesis.There is sufficient evidence to conclude that mean is not equal to 3500.

b)As,p-value=0.0000,so smallest level of significance will be 0 to be willing to reject the null hypothesis.

c)

t*=tinv(0.05,11)=2.201

margin of error=2.201*60/sqrt(12)=38.12

95% confidence interval for mean

=3250+/-38.12

=(3211.88, 3288.12)

As,above CI does not include 3500 within its limits,so we can conclude that mean is different from 3500.