Question

A generics manufacturer claims that the average tensile strength of its threadA is the same as its name brand competitors thread B. To test this claim, 50 pieces of each type ofthread were tested under similar conditions. Type A thread had an average tensile strength of 42.7kilograms while type B thread had an average tensile strength of 43.9 kilograms. If the populationstandard deviations of thread A and thread B are 7.35 and 6.29 respectively, can you conclude up to aα= 0.05level of significance that the tensile strength of thread A is less than thread B?

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 < μ2

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.645.
Hence reject H0 if z < -1.645

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(54.0225/50 + 39.5641/50)
sp = 1.3681

Test statistic,
z = (x1bar - x2bar)/sp
z = (42.7 - 43.9)/1.3681
z = -0.88

P-value Approach
P-value = 0.1894
As P-value >= 0.05, fail to reject null hypothesis.