Question

Tensile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data.

Grade Sample Size sample mean(kg/mm^2) Sample S
AISI 1064 m = 126 x = 102.8 s1 = 1.2
AISI 1078 n = 126 y = 121.3 s2 = 2.0

a. Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm^2.
b. Test the appropriate hypotheses using the P-value approach.
c. Calculate the test statistic and p-value.
d. State the conclusion in the problem context.

Answer 1

The null hypothesis is rejected.

There is enough evidence to support the claim that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm^2.

Test statistic t=-40.91

P-value = 0

Explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm^2.

Then, the null and alternative hypothesis are:

`H_0: \mu_1-\mu_2=-10\\\\H_a:\mu_1-\mu_2< -10`

The significance level is 0.05.

The sample 1 (AISI 1064), of size n1=126 has a mean of 102.8 and a standard deviation of 1.2.

The sample 2 (AISI 1078), of size n2=126 has a mean of 121.3 and a standard deviation of 2.

The difference between sample means is Md=-18.5.

`M_d=M_1-M_2=102.8-121.3=-18.5`

The estimated standard error of the difference between means is computed using the formula:

`s_(M_d)=\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}=\sqrt{(1.2^2)/(126)+(2^2)/(126)}\\\\\\s_(M_d)=√(0.011+0.032)=√(0.043)=0.2078`

Then, we can calculate the t-statistic as:

`t=(M_d-(\mu_1-\mu_2))/(s_(M_d))=(-18.5-(-10))/(0.2078)=(-8.5)/(0.2078)=-40.91`

The degrees of freedom for this test are:

`df=n_1+n_2-2=126+126-2=250`

This test is a left-tailed test, with 250 degrees of freedom and t=-40.91, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=P(t<-40.91)=0`

As the P-value (0) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm^2.