# Hypothesis Testing

WatchPage 1 of 1

Go to first unread

Skip to page:

Making a bit of a mess of my stats homework, so if someone could check I've got this question right then that would be great

In a local driving test centre the probability of passing your driving test for the first time is 54%. Mike claims that his driving school is better than this. He records the first time pass rate for 50 randomly selected learners he teaches.

1) At a 5% significance level, state your hypothesis clearly and find the critical region.

2) If 31 of Mike's students pass, state the conclusion Stan should make.

So for 1) The null hypothesis (or H0)=0.54

H1 > 0.54

By using the level of significance, the critical value is 34, and therefore the critical region would be >34

2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

Funnily enough now I've written it out this seems to make a lot more sense, but just in case I'm entirely on the wrong track it would be great if someone could confirm this is right

In a local driving test centre the probability of passing your driving test for the first time is 54%. Mike claims that his driving school is better than this. He records the first time pass rate for 50 randomly selected learners he teaches.

1) At a 5% significance level, state your hypothesis clearly and find the critical region.

2) If 31 of Mike's students pass, state the conclusion Stan should make.

So for 1) The null hypothesis (or H0)=0.54

H1 > 0.54

By using the level of significance, the critical value is 34, and therefore the critical region would be >34

2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

Funnily enough now I've written it out this seems to make a lot more sense, but just in case I'm entirely on the wrong track it would be great if someone could confirm this is right

0

reply

Report

#2

(Original post by

Making a bit of a mess of my stats homework, so if someone could check I've got this question right then that would be great

In a local driving test centre the probability of passing your driving test for the first time is 54%. Mike claims that his driving school is better than this. He records the first time pass rate for 50 randomly selected learners he teaches.

1) At a 5% significance level, state your hypothesis clearly and find the critical region.

2) If 31 of Mike's students pass, state the conclusion Stan should make.

So for 1) The null hypothesis (or H0)=0.54

H1 > 0.54

By using the level of significance, the critical value is 34, and therefore the critical region would be >34

2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

Funnily enough now I've written it out this seems to make a lot more sense, but just in case I'm entirely on the wrong track it would be great if someone could confirm this is right

**Lemur14**)Making a bit of a mess of my stats homework, so if someone could check I've got this question right then that would be great

In a local driving test centre the probability of passing your driving test for the first time is 54%. Mike claims that his driving school is better than this. He records the first time pass rate for 50 randomly selected learners he teaches.

1) At a 5% significance level, state your hypothesis clearly and find the critical region.

2) If 31 of Mike's students pass, state the conclusion Stan should make.

So for 1) The null hypothesis (or H0)=0.54

H1 > 0.54

By using the level of significance, the critical value is 34, and therefore the critical region would be >34

2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

Funnily enough now I've written it out this seems to make a lot more sense, but just in case I'm entirely on the wrong track it would be great if someone could confirm this is right

0

reply

Report

#3

**Lemur14**)

Making a bit of a mess of my stats homework, so if someone could check I've got this question right then that would be great

In a local driving test centre the probability of passing your driving test for the first time is 54%. Mike claims that his driving school is better than this. He records the first time pass rate for 50 randomly selected learners he teaches.

1) At a 5% significance level, state your hypothesis clearly and find the critical region.

2) If 31 of Mike's students pass, state the conclusion Stan should make.

So for 1) The null hypothesis (or H0)=0.54

H1 > 0.54

By using the level of significance, the critical value is 34, and therefore the critical region would be >34

2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

Funnily enough now I've written it out this seems to make a lot more sense, but just in case I'm entirely on the wrong track it would be great if someone could confirm this is right

The critical region is the region that will lead to the rejection of the null hypothesis. I disagree that this is > 34 (but it's close). Can you please post how you got that e.g. by using a calculator?

1

reply

(Original post by

Firstly 0.54 is not a hypothesis. Have you written the hypothesis properly and defined the random variable in your actual working? If not you should do that first and let us know if you need help.

The critical region is the region that will lead to the rejection of the null hypothesis. I disagree that this is > 34 (but it's close). Can you please post how you got that e.g. by using a calculator?

**Notnek**)Firstly 0.54 is not a hypothesis. Have you written the hypothesis properly and defined the random variable in your actual working? If not you should do that first and let us know if you need help.

The critical region is the region that will lead to the rejection of the null hypothesis. I disagree that this is > 34 (but it's close). Can you please post how you got that e.g. by using a calculator?

Haven't declared the random variable, whoops! Will add that now

Going well here...critical region should be

__>__34 I think?

Yeah it was using a calculator. I found that at 33 it was 5.8% and at 34 it was 3.1% so since the critical value is included in the critical region then the critical value would be 34 rather than 33.

(Original post by

You can check your answers here. http://burymathstutor.co.uk/binom.html

**BuryMathsTutor**)You can check your answers here. http://burymathstutor.co.uk/binom.html

1

reply

Report

#5

(Original post by

H0=0.54

Haven't declared the random variable, whoops! Will add that now

Going well here...critical region should be

Yeah it was using a calculator. I found that at 33 it was 5.8% and at 34 it was 3.1% so since the critical value is included in the critical region then the critical value would be 34 rather than 33.

While that calculator looks great, given I'm not confident in my hypothesis etc. then I needed someone would check I'm on the right track with the hypothesis etc. first rather than the answer calculator! Thanks anyway

**Lemur14**)H0=0.54

Haven't declared the random variable, whoops! Will add that now

Going well here...critical region should be

__>__34 I think?Yeah it was using a calculator. I found that at 33 it was 5.8% and at 34 it was 3.1% so since the critical value is included in the critical region then the critical value would be 34 rather than 33.

While that calculator looks great, given I'm not confident in my hypothesis etc. then I needed someone would check I'm on the right track with the hypothesis etc. first rather than the answer calculator! Thanks anyway

i.e.

and say what p is too.

0

reply

Thank you

0

reply

Report

#7

(Original post by

H0=0.54

Haven't declared the random variable, whoops! Will add that now

Going well here...critical region should be

Yeah it was using a calculator. I found that at 33 it was 5.8% and at 34 it was 3.1% so since the critical value is included in the critical region then the critical value would be 34 rather than 33.

**Lemur14**)H0=0.54

Haven't declared the random variable, whoops! Will add that now

Going well here...critical region should be

__>__34 I think?Yeah it was using a calculator. I found that at 33 it was 5.8% and at 34 it was 3.1% so since the critical value is included in the critical region then the critical value would be 34 rather than 33.

H0 = 0.54 still doesn't really make sense. First define your random variable X then write X~B(50, p).

The null hypothesis would then be : p = 0.54

and the alternate hypothesis is p > 0.54.

It's important you do all this correctly because you'll get marks for it in exams.

0

reply

Report

#8

(Original post by

2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

**Lemur14**)2) Hence, since 31 is not within the critical region, you accept the null hypothesis.

BuryMathsTutor is this correct or am I being nit picky?

0

reply

(Original post by

This is the correct critical region.

H0 = 0.54 still doesn't really make sense. First define your random variable X then write X~B(50, p).

The null hypothesis would then be : p = 0.54

and the alternate hypothesis is p > 0.54.

It's important you do all this correctly because you'll get marks for it in exams.

**Notnek**)This is the correct critical region.

H0 = 0.54 still doesn't really make sense. First define your random variable X then write X~B(50, p).

The null hypothesis would then be : p = 0.54

and the alternate hypothesis is p > 0.54.

It's important you do all this correctly because you'll get marks for it in exams.

Thank you

0

reply

Report

#10

(Original post by

Instead of saying you accept the null hypothesis I prefer to say that "there's no evidence to reject the null hypothesis". In a single hypothesis test your aim is to try to reject the null hypothesis, not provide evidence to back it up.

BuryMathsTutor is this correct or am I being nit picky?

**Notnek**)Instead of saying you accept the null hypothesis I prefer to say that "there's no evidence to reject the null hypothesis". In a single hypothesis test your aim is to try to reject the null hypothesis, not provide evidence to back it up.

BuryMathsTutor is this correct or am I being nit picky?

But I'm not the statistics expert here. Should we bother Gregorius?

0

reply

**Notnek**)

Instead of saying you accept the null hypothesis I prefer to say that "there's no evidence to reject the null hypothesis". In a single hypothesis test your aim is to try to reject the null hypothesis, not provide evidence to back it up.

BuryMathsTutor is this correct or am I being nit picky?

Thank you

0

reply

Report

#12

(Original post by

No, I think you're correct. A phrase I have read is "..there is insufficient evidence to reject the null hypothesis.."

But I'm not the statistics expert here. Should we bother Gregorius?

**BuryMathsTutor**)No, I think you're correct. A phrase I have read is "..there is insufficient evidence to reject the null hypothesis.."

But I'm not the statistics expert here. Should we bother Gregorius?

Gregorius For a simple hypothesis test, if the value of the random variable

*does not*fall within the critical region, would you say that a statement like "we accept the null hypothesis" is incorrect? This is compared to a statement like "there is insufficient evidence to reject the null hypothesis".

I always use the latter but I'm wondering if you would say it was

*incorrect*to say the former?

0

reply

Report

#13

(Original post by

Yes why not

**Notnek**)Yes why not

Gregorius For a simple hypothesis test, if the value of the random variable

I always use the latter but I'm wondering if you would say it was

*does not*fall within the critical region, would you say that a statement like "we accept the null hypothesis" is incorrect? This is compared to a statement like "there is insufficient evidence to reject the null hypothesis".I always use the latter but I'm wondering if you would say it was

*incorrect*to say the former?
4

reply

Report

#14

(Original post by

Yes, it would be incorrect, because saying that we "accept the null hypothesis" assumes that we have evidence in its favour, which need not be the case. Let me give you a toy example to illustrate. Suppose you set up a statistical test that has very little (or no) power to distinguish between the null and the alternative. Then your failure to reject the null is no evidence for the null hypothesis, but merely that you've got a lousy statistical test!

**Gregorius**)Yes, it would be incorrect, because saying that we "accept the null hypothesis" assumes that we have evidence in its favour, which need not be the case. Let me give you a toy example to illustrate. Suppose you set up a statistical test that has very little (or no) power to distinguish between the null and the alternative. Then your failure to reject the null is no evidence for the null hypothesis, but merely that you've got a lousy statistical test!

Your will is my command, oh master!

0

reply

Report

#15

(Original post by

Please return to your stats den. We'll let you know when we need you

**Notnek**)Please return to your stats den. We'll let you know when we need you

What you don't get in this framework is P(H1 | t) and/or P(H0 |t). To get these, you need to enter into a Bayesian framework - so start playing around with Bayes' formula...

A good example of a lousy statistic is t = 1, whatever the data. The P(t = 1 | H0) = P(t = 1 | H1) = 1. Then you'll simply get that P(H0 | t = 1) = P(H0), the prior probability of H0. As you might expect, your lousy statistic has told you nothing about the probability of H0 that you didn't already know.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top