Question

1) The sample mean and standard deviation from a random sample of 21 observations from a normal population were computed as ?¯=25 and s = 8. Calculate the t statistic of the test required to determine whether there is enough evidence to infer at the 6% significance level that the population mean is greater than 21.

Test Statistic =

2) 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 146 golfers be chosen so that their mean driving distance is 234.1 yards. The population standard deviation is 49.5 Use a 5% significance level.

Calculate the followings for a hypothesis test where ?0:?=235 and ?1:?<235

(a)    The test statistic is =

(b)    The P-Value is=

The final conclusion is

A. There is sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 235
B. There is not sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 235

Answer 1

1) Here claim is that mean is greater than 21

So hypothesis is vs

Test statistics is

So P value is TDIST(2.29,20,1)=0.017

As P value is less than , we reject the null hypothesis

Hence there is sufficient evidence to support the claim that mean is greater than 21

2) As population standard deviation is known, so we will use z statistics

a.

b. P value is

As P value is greater than , we fail to reject the null hypothesis

So answer here is

A. There is sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 235