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 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 = ______________

Answer 1

Answer:

(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: $H_0:\mu=235$

Alternative hypothesis: $H_1:\mu>235$

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

The formula for test statistic

$z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

where, z is test statistic, $\overline{x}$ is sample mean, $\sigma$ is population standard deviation and $\mu$ is population mean.

(a)

The value test statistics is

$z=\frac{240.4-235}{\frac{42.2}{\sqrt{130}}}$

$z=1.45899225013$

$z\approx 1.459$

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.

Answer 2

The standard normal distribution is normal distribution with mean 0 and standard deviation 1.

The hypothesis test conducted for given condition gives

• Test statistic: z = 1.46 approx.
• p value: 0.0721 approx

How to form the hypotheses?

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.

Null hypothesis is the one which researchers try to disprove.

For the given case, we have the hypotheses as

Null hypothesis : $H_0 : \mu = 235$

Alternate hypothesis : $H_1 : \mu > 235 \text{\: (Single tailed test)}$

It is given that:

Sample size: n = 130

Sample mean by null hypothesis: $\overline{x} = 235$ yards

Population standard deviation: $\sigma =42.2$ yards

The test statistic is calculated as

$z = \dfrac{\overline{x} - \mu}{\sigma / \sqrt{n}} = \dfrac{240.4 - 235}{ 42.2/\sqrt{130}}\\\\z \approx 1.46$

Its p value from the z tables is obtained as 0.9279.

This is not the intended p value as it is showing the left sided area to the z score 1.46. We need right side area as the test is right tailed.

As total area is 1 for standard normal distribution, thus, we get:

p value for right tailed test at z score 1.46 = 1 - 0.9279 = 0.0721

Thus, The hypothesis test conducted for given condition gives

• Test statistic: z = 1.46 approx.
• p value: 0.0721 approx

Learn more about hypothesis testing here:
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