Random Sample Size = n =101
Mean driving distance = Sample Mean = = 239.6
Population Standard Deviation = = 43.5
Level of significance () = 5% = 0.05
The hypothesis for the test is given as-
a) The test statistic Z is given as -
Therefore value of test statistic = -0.09
b)
Therefore p value = 0.4641
We can see that our p value is greater than level of significance (), so we will accept null hypothesis.
A. There is sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 240.
Hence the solution.
(6 points) Golf-course designers have become concerned that old courses are becoming obsolete since new technology...
(6 points) 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 240 yards on average. Suppose a random sample of 101 golfers be chosen so that their mean driving distance is 239.6 yards. The population standard deviation is 43.5. Use a...
(6 points) 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 250 yards on average. Suppose a random sample of 140 golfers be chosen so that their mean driving distance is 248 yards. The population standard deviation is 44.7. Use a...
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 250 yards on average. Suppose a random sample of 198 golfers be chosen so that their mean driving distance is 248.8 yards. The population standard deviation is 47.3. Use a 5% significance...
(1 point) 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 151 golfers be chosen so that their mean driving distance is 234.7 yards. The population standard deviation is 40.9. Use...
Question 6 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. A random sample of 177 golfers show that their mean driving distance is 230.7 yards, with a standard deviation of 41.8. Set up the null and...
that new golf courses need to be buit expecting that the average gofer can hit the ball more than 240 yards on average. Suppose a random sample of 175 golers be chosen of 41.2. so that their mean driving distance is 241.6 yards, with a standard deviation (a) test statistic You have 3 attempts remaining.