Perform the calculations required to respond to the following business problem u

Question

Perform the calculations required to respond to the following business problem using the appropriate Excel hypothesis testing spreadsheets which can be found on Blackboard. Calculations done using any other method will not be accepted and a zero will be recorded for your score. Use an alpha value of.05 on this problem Studies conducted by a manufacturer of Boston and Vermont brand asphalt shingles have shown product weight to be a major factor in customers' perception of quality. Moreover, the weight represents the amount of raw material being used and is therefore vary important to the company from a cost standpoint. The last stage of the assembly line packages the shingles before the packages are placed on wooden pallets. When a pallet is full (a pallet for most brands holds 16 squares of shingles), it is weighed, and the measurement is recorded. The data file Pallet (found on Blackboard in the Excel Files folder) contains the weight (in pounds) from a sample of 368 pallets of Boston brand shingles and 330 pallets of Vermont brand shingles. Use the Excel statistical functions AVERAGE and STDEV.S to calculate the mean and standard deviation of the data in the two files. For the Boston brand shingles, is there evidence that the population mean weight is different from 3,150 pounds? For the Vermont brand shingles, is there evidence that the population mean weight is different from 3,700 pounds? Print the resulting Excel worksheets and write a conclusion statement at the bottom of eaclh sheet. Hand in these worksheets at the start of the exam.

Explanation / Answer

For the Boston brand shingles, is there evidence that the population mean weight is different from 3150 pounds?

Solution:

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

H0: µ = 3150 versus Ha: µ 3150 (Two tailed test)

We are given

Level of significance = = 0.05

Sample size = n = 368

Sample mean = Xbar = 3124.21

Sample standard deviation = S = 34.71

Degrees of freedom = n – 1 = 367

Lower critical value = -1.9664

Upper critical value = 1.9664

Test statistic = t = (Xbar - µ)/[S/sqrt(n)]

Test statistic = t = (3124.21 – 3150)/[34.71/sqrt(368)]

Test statistic = t = -14.2535

P-value = 0.00

(Critical value and P-value are calculated by using t-table)

P-value < Level of significance =

So, we reject the null hypothesis

There is sufficient evidence to conclude that population mean weight is different from 3150 pounds.

For the Vermont brand shingles, is there evidence that the population mean weight is different from 3700 pounds?

Solution:

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

H0: µ = 3700 versus Ha: µ 3700 (Two tailed test)

We are given

Level of significance = = 0.05

Sample size = n = 330

Sample mean = Xbar = 3704.04

Sample standard deviation = S = 46.74

Degrees of freedom = n – 1 = 329

Lower critical value = -1.9672

Upper critical value = 1.9672

Test statistic = t = (Xbar - µ)/[S/sqrt(n)]

Test statistic = t = (3704.04 – 3700)/[46.74/sqrt(330)]

Test statistic = t = 1.5702

P-value = 0.1173

(Critical value and P-value are calculated by using t-table)

P-value > Level of significance =

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that population mean weight is different from 3700 pounds.

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