Question

In: Statistics and Probability

Driving Force is a golf ball manufacturer. Their R & D department have been developing a...

Driving Force is a golf ball manufacturer. Their R & D department have been developing a type of golf ball with a new dimpling pattern that is designed to increase the flight distance of the ball. They are at the testing stage and run an experiment to test the mean flight distance for the new type of ball compared to a standard one. A special device is used to fire the balls at a set force and angle.

A sample of 30 of the new balls (sample A) and 30 standard balls (sample B) are fired using the device and the distance travelled (in feet) is recorded for each ball.

Download the data

Sample A
798 784 778 808 794
802 767 768 835 827
790 675 765 737 774
810 787 787 746 768
785 707 799 755 732
768 737 808 830 780
Sample B
715 778 814 739 717
813 776 778 766 774
758 716 785 761 761
699 758 773 788 713
704 755 759 818 776
676 790 795 755 754

Conduct an appropriate one-tailed hypothesis test. Assume that the population variances are not equal.

a)From the following options, select the correct null and alternate hypotheses for this test:

A:H0: μA < μB, Ha: μA > μB

B:H0: μA = μB, Ha: μA < μB

C:H0: μA = μB, Ha: μA > μB

D:H0: μA = μB, Ha: μA ≠ μB

The correct null and alternate hypotheses for this test are: ABCD

b)Calculate the test statistic. Give your answer to 3 decimal places.

t =

c)At a significance level of 0.05, the null hypothesis is rejectednot rejected.

That is, you can state that there is proofsignificant evidencenot enough evidence to conclude that the population mean flight distancesample mean flight distance of the new type of golf ball is less thangreater thanequal to that of the standard golf ball.

Solutions

Expert Solution

For sample A

= 776.7, s1^2 = 35.468

For sample B

= 758.8, s2 = 35.531

a) H0:

    H1:

b) The test statistic t = ()/sqrt(s12/n1 + s2^2/n2)

                               = (776.7 - 758.8)/((35.468)^2/30 + (35.531)^2/30)

                             = 0.213

At alpha = 0.05, the critical value is t* = -1.699

Since the test statistic value is not less than critical value, so we should not reject the null hypothesis.

c) = At a significance level of0.05 the null hypothesis is not rejected.

That is, we can state that there is not enough evidence to conclude that the population mean flight distance is less than to the standard golf ball.


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