Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 230 yards on average. suppose a random sample of 152 golfers be chosen so that their mean driving distance is 229.9 yards. the population standard deviation is 41.4. use a 5% significance level.

Complete Question:

Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 230 yards on average. suppose a random sample of 152 golfers be chosen so that their mean driving distance is 229.9 yards. the population standard deviation is 41.4. use a 5% significance level.

Calculate the followings for a hypothesis test where H0:μ = 230 and H1:μ < 230 :

(a) The test statistic is

(b) The P-Value is

The final conclusion is:

t - statistic = -0.02977977

p value = 0.488753

There is no sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 230

Explanation:

Number of samples(n) = 152

Significance level = 5%

sample standard deviation(s) = 41.4

sample Mean (x) = 229.9

H0:μ = 230 and H1:μ < 230

A.) The test statistic can be calculated using the formula;

t = (x - μ) ÷ (s/sqrt(n))

t = (229.9 - 230) ÷ (41.4/sqrt(152))

t = -0.1 ÷ 3.3579834174

t-statistic = -0.02977977

Using the p value calculator on the t-statistic value at a significance level of 0.05;

The p-value is 0.488753

With the p value sufficiently greater than 0.05, There is no sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 230

im going to come back to you?