A manufacturer claims that the mean tensile strength of thread A exceeds the average tensile strength of thread B. To test his claim, 16 sample pieces of each type of thread are tested under similar conditions. Type A thread had a sample average tensile strength of 185 kg with a standard deviation of 6 kg, while type B thread had a sample average tensile strength of 178 kg with a standard of 9 kg. Assume that both populations are normally distributed and the variances are equal. Test the manufacturers claim using a = 0.05 level of significance.

The complete part of the first sentence is;

A manufacturer claims that the average tensile strength of thread A exceeds the average tensile strength of thread B by at least 12 kilograms.

we fail to reject the null hypothesis and conclude that the difference of the average tensile strength of thread A and thread B is less than 12

Step-by-step explanation:

We are given;

n_A = 16

n_B = 16

x'_A = 185 kg

x'_B = 178 kg

s_A = 6 kg

s_B = 9 kg

Let μ_A denote the population average tensile strength of thread A

Also, Let μ_B represent the population average tensile strength of thread B

Thus;

Null Hypothesis; H0;μ_A - μ_B ≤ 12

Alternative hypothesis;H1; μ_A - μ_B > 12

From the image attached, with a significance level of 0.05, the critical value for right tailed is 1.645. So we will reject the hypothesis is z > 1.645

Formula for z is;

z = (x'_A - x'_B - d_o)/√((s_A²/n_A) + (s_B²/n_B))

Plugging in the relevant values, we have;

z = (185 - 178 - 12)/√((6²/16) + (9²/16))

z = -5/2.7041634566

z = - 1.849

Since the z-value is less than 1.645,we fail to reject the null hypothesis and conclude that the difference of the average tensile strength of thread A and thread B is less than 12

20335+282=20617

step-by-step explanation:

do you have a picture of the table