Q:

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 235 yards on average. Suppose a random sample of 130 golfers be chosen so that their mean driving distance is 240.4 yards, with a population standard deviation of 42.2 .Conduct a hypothesis test where H_0: \mu = 235 and H_1:\mu > 235 by computing the following:(a) \ test statistic ______________\ (b) \ p-value p = ______________

Accepted Solution

A:
Answer:a) e) H0 : µ ≤ 230 versus Ha : µ > 230b) [tex]t=\frac{230.7-230}{\frac{41.8}{\sqrt{177}}}=0.223[/tex]    c. 0.223 c) [tex]p_v =P(t_{(176)}>0.223)=0.4118[/tex]  d. 0.4118 Step-by-step explanation:Information given  [tex]\bar X=230.7[/tex] represent the sample mean[tex]s=41.8[/tex] represent the sample standard deviation [tex]n=177[/tex] sample size  [tex]\mu_o =230[/tex] represent the value to verifyt would represent the statistic  [tex]p_v[/tex] represent the p valuePart aWe want to verify if the true mean is higher than 230 yards, the system of hypothesis would be:  Null hypothesis:[tex]\mu \leq 230[/tex]  Alternative hypothesis:[tex]\mu > 230[/tex]  The best option would be:H0 : µ ≤ 230 versus Ha : µ > 230Part bThe statistic is given by:[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  Replacing the info given we got:[tex]t=\frac{230.7-230}{\frac{41.8}{\sqrt{177}}}=0.223[/tex]    Part cThe degrees of freedom are:[tex]df=n-1=177-1=176[/tex]  The p value would be given by:[tex]p_v =P(t_{(176)}>0.223)=0.4118[/tex]  d. 0.4118