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.
Answer:
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
