Question

6. The thickness of six pads designed for use in aircraft engine mounts were measured. The results, in mm, were: 40.93, 41.11, 41.47, 40.96, 40.80, and 41.32. Assume that these values come from a normal population.

   1. (a) Find the sample mean and the sample variance.

   2. (b) The target thickness is 41.2 mm. Can you conclude that the mean thickness differs from the target value at 0.05 level of significance?

   3. (c) Find the P -value in (b), and interpret the result.

Answer 1

(a) Find the sample mean and the sample variance.

Answer:

From given data, we have

Sample mean = Xbar = 41.09833333

Sample standard deviation = S = 0.254512606

Sample variance = S^2 = 0.254512606^2 = 0.064777

(Mean and SD are calculated by using excel)

(b) The target thickness is 41.2 mm. Can you conclude that the mean thickness differs from the target value at 0.05 level of significance?

Solution:

Here, we have to use one sample t test for the population mean.

H₀: µ = 41.2 versus H_a: µ ≠ 41.2

WE are given α = 0.05

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

We are given

Xbar = 41.09833333

S = 0.254512606

n = 6

µ = 41.2

df = n – 1 = 6 – 1 = 5

Critical values = - 2.5706 and 2.5706

(by using t-table)

t = (Xbar - µ)/[S/sqrt(n)]

t = (41.09833333 – 41.2)/[ 0.254512606/sqrt(6)]

t = -0.9785

Test statistic = t = -0.9785

Test statistic value is lies within Critical values = - 2.5706 and 2.5706

So, we do not reject the null hypothesis

We cannot conclude that the mean thickness differs from the target value at 0.05 level of significance.

(c) Find the P -value in (b), and interpret the result.

P-value = 0.3728

(by using t-table)

α = 0.05

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the mean thickness differs from the target value at 0.05 level of significance.