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 =

(a) The test statistics is z=1.459.

(b) The P-value is 0.072283.

Step-by-step explanation:

Given information:

Sample size = 130 golfers

Population standard deviation = 42.2

Sample mean = 240.4 yards

Population mean = 235 yards

Assume that the data is normally distributed.

Null hypothesis:

Alternative hypothesis:

It is a right tailed test and significance level is α=0.05.

The formula for test statistic

where, z is test statistic, is sample mean, is population standard deviation and is population mean.

(a)

The value test statistics is

Therefore the test statistics is z=1.459.

(b)

The P-value for left tailed test statistics z=1.459 at signification level α=0.05 is 0.072283.

Therefore the P-value is 0.072283.

Since p-value > α, therefore we accept the null hypothesis.

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles

answer:

125 yards

explanation:

75+50=125. that is my answer