Question

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 784 863 735 759 813 791 775 814 796 785 795 775 832 790 751 785 734 759 772 800 803 777 736 849 729 787 804 802 800 736

Sample B 719 761 764 845 723 759 815 728 789 786 791 819 774 737 801 754 797 807 707 799 811 779 752 785 824 760 739 781 796 764

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: μ_A = μ_B, H_a: μ_A < μ_B

B:H₀: μ_A = μ_B, H_a: μ_A > μ_B

C:H₀: μ_A < μ_B, H_a: μ_A > μ_B

D:H₀: μ_A = μ_B, H_a: μ_A ≠ μ_B

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 tothat of the standard golf ball.

Answer 1

Solution:

Part a)

Here, we have to use two sample t test for the population means assuming unequal population variances.

The null and alternative hypotheses for this test are given as below:

H₀: µ_A = µ_B versus H_a: µ_A > µ_B

This is a one tailed test. (Upper tailed/Right tailed)

Part b)

We are given α = 0.05

Two sample t test for mean assuming unequal population variances

t = (X1bar – X2bar) / sqrt[(S₁² / n₁)+(S₂² / n₂)]

From given data, we have

X1bar = 784.3666667

X2bar = 775.5333333

S1 = 33.0584

S2 = 33.9134

n1 = 30

n2 = 30

df = Min. (n1 – 1, n2 – 1) = 29

t = (784.3666667 - 775.5333333) / sqrt[(33.0584^2/30)+( 33.9134^2/30)]

t = 8.8333/8.6467

t = 1.0216

Answer: t = 1.022

Part c

From above test statistic value and by using t-table or excel, we have an approximate p-value given as below:

P-value = 0.1577

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

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 population mean flight distance of the new type of golf ball is greater than that of the standard golf ball.