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# stats

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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 255 yards on average. Suppose a random sample of 166 golfers be chosen so that their mean driving distance is 259.5 yards, with a standard deviation of 44. Conduct a hypothesis test where H0:μ=255 and H1:μ>255 by computing the following: (a) test statistic (b) p-value p= (c) If this was a two-tailed test, then the p-value is

Apr 4, 2020

#1
#2
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Cal you have sent this user to a site where you have to pay to see the answer.

I do not encourage you to do that. Not without telling the asker in advance anyway.

Just be a little conscious that some sites will not provide an answer without payment first.

Melody  Apr 4, 2020
#7
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Oops....... Sorry. This won't happen again!

CalTheGreat  Apr 4, 2020
#3
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Here is a video on this.

If you do not like this video there are plenty of others.

Apr 4, 2020
edited by Melody  Apr 4, 2020
#4
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did you understand the video??

Guest Apr 4, 2020
#5
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Are you the original guest or a new guest?

Melody  Apr 4, 2020
edited by Melody  Apr 4, 2020
#6
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i was asking you, im a new guest

Guest Apr 4, 2020
#9
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Melody did you watch the video?

Guest Apr 4, 2020
#8
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Melody did you watch the video

Apr 4, 2020
#15
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Why did my answer get flagged

Apr 4, 2020
#16
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idk but it is fixed now.

Melody  Apr 4, 2020
#17
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Yes I did watch the video and Yes i did understand the video.

Apr 4, 2020
#18
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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 255 yards on average. Suppose a random sample of 166 golfers be chosen so that their mean driving distance is 259.5 yards, with a standard deviation of 44. Conduct a hypothesis test where H0:μ=255 and H1:μ>255 by computing the following: (a) test statistic (b) p-value p= (c) If this was a two-tailed test, then the p-value is

$$\mu=255\\ n=166\\ \bar X=259.5\\ S=44\\ Test \;Stat \;is \;Z\\ Z=\frac{\bar X-\mu}{\left( \frac{S}{\sqrt n} \right)}\\ Z=\frac{259.5-255}{\left( \frac{44}{\sqrt {166}} \right)}\\ Z=1.318\\ \\~\\ H_0:\quad \mu=255\\ H_A:\quad \mu>2.55\\ \\~\\ P(Z>1.318) =P(Z<-1.38) \qquad \text{You can use either}$$

Maybe you have to use a given table.

I like this site.

http://davidmlane.com/hyperstat/z_table.html

$$\boxed{\bf\text {It gives the P value as 0.0938}}$$ I am reasonably sure that is all correct.

Coding

\mu=255\\
n=166\\
\bar X=259.5\\
S=44\\
Test \;Stat \;is \;Z\\
Z=\frac{\bar X-\mu}{\left(    \frac{S}{\sqrt n}   \right)}\\
Z=\frac{259.5-255}{\left(    \frac{44}{\sqrt {166}}   \right)}\\
Z=1.318\\
\\~\\

\\~\\
P(Z>1.318)
=P(Z<-1.38)

Apr 4, 2020
edited by Melody  Apr 4, 2020
#19
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If it is a 2 tail test I do not know what the answer is. I'd have to watch the next video I guess.

Apr 4, 2020
#20
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are you familiar with p values

Guest Apr 4, 2020
#21
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Why don't you just say what you want to say in plain English?

Better still if you believe you are knowledgable why don't you just answer the question yourself.

Why are you hiding behind anonymity?

People who make snide comments behind a curtain of anonymity usually have no idea of any relevant thing.

Melody  Apr 4, 2020