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:
Answer:
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
