tables: Sample A 828 812 788 846 785 790 798 786 742 826 739 817 780 803 791 809

Question

tables:

Sample A 828 812 788 846 785 790 798 786 742 826 739 817 780 803 791 809 746 746 705 742 780 771 799 736 728 742 800 775 824 755 734 826 749 789 770 785 770 747 828 740 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 40 of the new balls (sample A) and 40 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 Sample B 788 785 718 788 783 780 730 828 812 846 790 798 786 742 826 761 818 759 803 755 739 817 780 803 791 801 791 692 681 709 809 746 746 705 742 740 724 792 768 727 780 771 799 736 728 806 768 775 766 790 748 802 792 780 735 742 800 775 824 755 734 826 749 789 770 693 734 779 821 753 785 770 747 828 740 784 796 775 802 760 Conduct a one-tailed hypothesis test for equality of the population means. Assume that the population variances are not equal a) From the following options, select the correct nu and alternate hypotheses for this test: HA HB HB HA HB HA C: H HB HA HB o: JA HBr HA NB The correct null and alternate hypotheses for this test are: b) Calculate the test statistic. Give your answer to 3 decimal places. c) At a significance level of 0.05, the n hypothesis is That is, you can state that there is to conclude that the of the new type of golf ba that of the standard golf ba

Explanation / Answer

(A)

Below are the null and alternate hypothesis for a one tailed test
H0: uA = uB
H1: uA < uB
Option (D)

(B)

t = 1.626

(C)

The P-Value is 0.053992.


At a significance level of 0.05, the null hypothesis is not rejected.
That is, you can state that there is not enough evidence to conclude that the sample mean flight differece of the new type of golf ball is greater than that of the standard golf ball.

x1(bar) 776.9 x2(bar) 764.48 s1 33.2202 s2 35.0899 n1 40 n2 40 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 27.5895 (s22/n2) 30.7825 SE 7.64 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 19.517509 [ (s22 / n2)2 / (n2 - 1) ] 24.296512 (s12/n1 + s22/n2)2 3407.2985 DF = 78

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