Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if it has greater length off the tee than the best-selling driver. To do this, they have selected a group of golfers of differing abilities and ages. Their plan is to have each player use each of the two clubs and hit five balls. They will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players are:new club leading club236.4 237.2202.5 200.4245.6 240.8257.4 259.3223.5 218.9205.3 200.6266.7 258.9240 236.5278.9 280.5211.4 206.5a. At the 0.05 level of significance, can the company conclude that the New Club has increased average driving distance compared with the Leading Club? show how you came to the conclusion

From the hypothesis test carried out, the p-value obtained < the significant level at which the test was performed at, hence, the company can conclude that the New Club has increased average driving distance compared with the Leading Club.

Step-by-step explanation:

The results of drives of 10 players using the new and the leading club is presented thus

New club | Leading club

236.4 | 237.2

202.5 | 200.4

245.6 | 240.8

257.4 | 259.3

223.5 | 218.9

205.3 | 200.6

266.7 | 258.9

240.0 | 236.5

278.9 | 280.5

211.4 | 206.5

The test is to check if the New Club has increased average driving distance compared with the Leading Club.

Since the two sets of average drives belong to the same set of people, a t-test is appropriate.

To do this test, we first take the difference in these average drives for New Club and Leading Club.

New club | Leading club | Difference

236.4 | 237.2 | -0.8

202.5 | 200.4 | 2.1

245.6 | 240.8 | 4.8

257.4 | 259.3 | -1.9

223.5 | 218.9 | 4.6

205.3 | 200.6 | 4.7

266.7 | 258.9 | 7.8

240.0 | 236.5 | 3.5

278.9 | 280.5 | -1.6

211.4 | 206.5 | 4.9

The mean of differences = (sum of the differences) ÷ (sample size)

Sum of the differences = -0.8 + 2.1 + 4.8 + -1.9 + 4.6 + 4.7 + 7.8 + 3.5 + -1.6 + 4.9 = 28.1

Sample size = 10

mean = (28.1/10) = 2.81

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = 13.0321+0.5041+3.9601+22.1841+3.2041+3.5721+24.9001+0.4761+19.4481+4.3681

= 95.649

σ = √[Σ(x - xbar)²/N] = √(95.649/10) = 3.093

To do an hypothesis test, we need to first define the null and alternative hypothesis

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

The test is to check if the New Club has increased average driving distance compared with the Leading Club.

The null hypothesis would then be that there is no significant evidence that the New Club has increased average driving distance compared with the Leading Club.

That is, the average driving distance with the new club isn't higher than the average driving distance with the leading club or the new club's average drive is lesser or equal to the average drive with the leading club or the difference of the average driving distance with the new club and the average driving distance with the leading club is less than or equal to 0.

The alternative hypothesis in the other hand would be that there is significant evidence that the New Club has increased average driving distance compared with the Leading Club.

That is, the difference of the average driving distance with the new club and the average driving distance with the leading club is greater than 0.

If the population difference of the average driving distance with the new club and the average driving distance with the leading club is μ

Mathematically,

The null hypothesis is represented as

H₀: μ ≤ 0

The alternative hypothesis is given as

Hₐ: μ > 0

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = mean difference between two sets of data obtained from the sample = 2.81

μ₀ = 0

σₓ = standard error = (σ/√n)

where n = Sample size = 10

σₓ = (σ/√n) = (3.093/√10) = 0.978

t = (2.81 - 0) ÷ 0.978

t = 2.87

checking the tables for the p-value of this t-statistic

Degree of freedom = df = 10 - 1 = 10 - 1 = 9

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing only in one direction. (Greater than 0 direction)

p-value (for t = 2.87, at 0.05 significance level, df = 9, with a one tailed condition) = 0.009238

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.009238

0.009238 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that the difference of the average driving distance with the new club and the average driving distance with the leading club is greater than 0.

Hope this Helps!!!

for the points

step-by-step explanation:

i think the answer is b.

step-by-step explanation:

not 100% sure though.