Question

The length of a construction part manufactured by a supplier follows a normal distribution with an unknown σ. The design length is 35.00 centimeters. Seven parts were randomly selected from the warehouse and measured. The actual lengths in centimeters were as follows: Part 1: 34.46 Part 2: 37.17 Part 3: 38.03 Part 4: 39.91 Part 5: 34.86 Part 6: 35.41 Part 7: 40.43

(a)[7] At α = 0.04, test to see if a typical manufactured part would conform to the design length. Make sure to define the hypotheses first, make a sketch of the test, and then carry out the test. (b)[3] Sketch and find the p‐value of the test. Would you reject the null if α = 0.07. Hint: Use 5 decimal places. Use some Excel lookups for values and probabilities.

Answer 1

Let be the true average length of the part. We want to test to see if a typical manufactured part would conform to the design length, which is 35 cms. That is, we want to test if .

a) The hypotheses are

Using the sample data, we know the following

n=7 is the sample size

The sample mean length is

The sample standard deviation is

We estimate the population standard deviation using the sample

The estimated standard error of mean is

The sample size n=7 is less than 30 and we do not know the population standard deviation. Since the population follows a normal distribution, we can say that the sampling distribution of means is normal distribution.

That is, we will do a 1 sample t test for means

The hypothesized value of the mean length is

The test statistic is

This is a 2 tailed test (The alternative hypothesis has "not equal to")

The right tail critical value for is

The degrees of freedom are n-1=7-1=6

Using the Excel function =T.INV.2T(0.04,6), we get 2.61224 (rounding to 5 decimals)

The critical values are -2.61224, +2.61224

The critical regions are <-2.61224 and >+2.61224. That is, we reject the null hypothesis, if the tets statistic lies in the critical region (outside the interval [-2.61224, +2.61224])

The sketch of the test is below

Here, we can see that the test statistic 2.40162 lies in the acceptance region. Hence we do not reject the null hypothesis.

ans: Do not reject H₀. There is no sufficient evidence to reject the claim that the mean length of the part is 35cm. We conclude that a typical manufactured part would conform to the design length.

b) This is a 2 tailed test. The p-value is the sum of the areas under both the tails

Using the Excel function =T.DIST.2T(2.40162,6), we get p-value=0.0532

We will reject the null hypothesis, if the p-value is less than the significance level.

As shown in the following sketch,

The p-value is 0.0532 and it is not less than alpha=0.04. Hence we do not reject the null hypothesis, when alpha=0.04.

ans: The p-value is 0.0532 and it is less than alpha=0.07. Hence we reject the null hypothesis, when alpha=0.07.