In: Statistics and Probability
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
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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.
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.