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Chapter 10: Q. 10.99 (page 430)

An independent golf equipment testing facility compared the difference in the performance of golf balls hit off a regular \(2-3/4"\) wooden tee to those hit off a \(3"\) Stinger Competition golf tee. A Callaway great Big Bertha driver with \(10\) degrees of loft was used for the test and a robot swung the club head at approximately \(95\) miles per hour. Data on ball velocity (in miles per hour) with each type of tee based on the test results are provided on Th WeiessStats site. Use the technology of your choice to do the following.

a. Obtain normal probability plots, boxplots and the standard deviation for the two samples.

b. Bases on your results from part (a) which would you be inclined to use to compare the population means: a pooled or a nonpooled \(t-\)procedure? Explain your answer.

c. At the \(5%\) significance level, do the data provide sufficient evidence to conclude that on average ball velocity is less with the regular tee than with the Stinger tee? Perform the required hypothesis test by using both the pooled \(t-\)test and the nonpooled \(t-\)test and compare results.

d. Find a \(90%\) confidence interval for the difference between the mean ball velocities with the regular and Stinger tees. Obtain the reuiqred confidence interval by using both the poold \(t-\)interval procedure and the nonpooled \(t-\)interval procedure. Compare your results.

Short Answer

Expert verified

Part a.

Part b. From the summary statistics of two samples, the pooled test is in appropriate because samples are considerably different.

Part c. The \(P(0.000)\leq (\alpha =0.05)\) so, the null hypothesis is rejected.

Part d. See Explanation

Step by step solution

01

Part a. Step 1. Given information

Consider the data,

02

Part a. Step 2. Calculation

The normal probability curve is,

And

The box plot is,

And

03

Part b. Step 1. Explanation

From the summary statistics of two samples, the pooled test is in appropriate because samples are considerably different.

04

Part c. Step 1. Calculation

For non pooled test,

The \(P(0.000)\leq (\alpha =0.05)\) so, the null hypothesis is rejected.

And for pooled,

The \(P(0.000)\leq (\alpha =0.05)\) so, the null hypothesis is rejected.

05

Part d. Step 1. Explanation

For non pooled test,

And for pooled test,

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Most popular questions from this chapter

A variable of two population has a mean of 7.9and standard deviation of 5.4for one of the population and a mean of 7.1and a standard deviation of 4.6 for the other population.

a. For independent samples of sizes 3and6respectively find the mean and standard deviation of x1-x2

b. Must the variable under consideration be normally distributed on each of the two population for you to answer part (a) ? Explain your answer.

b. Can you conclude that the variable x1-x2is normally distributed? Explain your answer.

Cooling Down. Cooling down with a cold drink before exercise in the heat is believed to help an athlete perform. Researcher 1. Dugas explored the difference between cooling down with an ice slurry (slushy) and with cold water in the article "lce Slurry Ingestion Increases Running Time in the Heat" (Clinical Journal of Sports Medicine, Vol. 21, No, 6, pp. 541-542). Ten male participants drank a flavored ice slurry and ran on a treadmill in a controlled hot and humid environment. Days later, the same participants drank cold water and ran on a treadmill in the same bot and humid environment. The following table shows the times, in minutes, it took to fatigue on the treadmill for both the ice slurry and the cold water.

At the 1%significance level, do the data provide sufficient evidence to conclude that, on average, cold water is less effective than ice slurry For optimizing athletic performance in the heat? (Note; The mean and standard deviation of the paired differences are -5.9minutes and 1.60minutes, respectively.)

In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from two populations. In each case, use the non pooled t-fest and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

x~1=20,s1=2,n1=30,x2=18,s2=5,n2=40.

a. Right-tailed test, α=0.05

b. 90%confidence interval.

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference,μ1-μ2, between two population means. interpret each confidence interval.

95% CI is from15to20

Discuss the basic strategy for comparing the means of two populations based on a simple random paired sample.
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