Null Hypothesis
Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of 80 dollars. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was %76 and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level.
State the null and alternative hypotheses for this test
Here is the t-distribution table
probability statistics
asked Apr 21 '14 at 21:41
Hani Abdullah
186
add a comment |
Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of 80 dollars. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was %76 and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level.
State the null and alternative hypotheses for this test
Here is the t-distribution table
probability statistics
asked Apr 21 '14 at 21:41
Hani Abdullah
186
1
Do you have a question of your own, or are you passing on to us a question written by someone other than yourself by just copying it? Phrasing a question in a manner suitable for assigning homework is frowned upon here.
– Michael Hardy
Apr 21 '14 at 21:55
No sir this not like that. I did solve that question with the help of my teacher but i need to know the exact procedure of it. So could u please help me out?
– Hani Abdullah
Apr 21 '14 at 22:18
add a comment |
Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of 80 dollars. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was %76 and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level.
State the null and alternative hypotheses for this test
Here is the t-distribution table
probability statistics
asked Apr 21 '14 at 21:41
Hani Abdullah
186
Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of 80 dollars. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was %76 and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level.
State the null and alternative hypotheses for this test
Here is the t-distribution table
probability statistics
probability statistics
asked Apr 21 '14 at 21:41
Hani Abdullah
186
asked Apr 21 '14 at 21:41
Hani Abdullah
186
edited Apr 23 '14 at 18:06
asked Apr 21 '14 at 21:41
Hani Abdullah
186
asked Apr 21 '14 at 21:41
Hani Abdullah
186
asked Apr 21 '14 at 21:41
Hani Abdullah
186
186
1
Do you have a question of your own, or are you passing on to us a question written by someone other than yourself by just copying it? Phrasing a question in a manner suitable for assigning homework is frowned upon here.
– Michael Hardy
Apr 21 '14 at 21:55
No sir this not like that. I did solve that question with the help of my teacher but i need to know the exact procedure of it. So could u please help me out?
– Hani Abdullah
Apr 21 '14 at 22:18
add a comment |
1
Do you have a question of your own, or are you passing on to us a question written by someone other than yourself by just copying it? Phrasing a question in a manner suitable for assigning homework is frowned upon here.
– Michael Hardy
Apr 21 '14 at 21:55
No sir this not like that. I did solve that question with the help of my teacher but i need to know the exact procedure of it. So could u please help me out?
– Hani Abdullah
Apr 21 '14 at 22:18
1
1
Do you have a question of your own, or are you passing on to us a question written by someone other than yourself by just copying it? Phrasing a question in a manner suitable for assigning homework is frowned upon here.
– Michael Hardy
Apr 21 '14 at 21:55
Do you have a question of your own, or are you passing on to us a question written by someone other than yourself by just copying it? Phrasing a question in a manner suitable for assigning homework is frowned upon here.
– Michael Hardy
Apr 21 '14 at 21:55
No sir this not like that. I did solve that question with the help of my teacher but i need to know the exact procedure of it. So could u please help me out?
– Hani Abdullah
Apr 21 '14 at 22:18
No sir this not like that. I did solve that question with the help of my teacher but i need to know the exact procedure of it. So could u please help me out?
– Hani Abdullah
Apr 21 '14 at 22:18
add a comment |
1 Answer
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active
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votes
I think instead of saying "the average cost to process a claim is approximately normally distributed with a mean of 80 dollars" you should have said "the cost to process a claim is approximately normally distributed with a mean of 80 dollars", letting the word "mean" be the only reference to averages.
The null hypothesis would be that the cost-cutting measures had no effect, so that that mean is still at least 80 dollars. The alternative hypothesis were effective, so that the mean is now lower than that.
Generally a null hypothesis is something presumed true until statistical evidence indicates otherwise.
answered Apr 22 '14 at 1:35
Michael Hardy
1
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think instead of saying "the average cost to process a claim is approximately normally distributed with a mean of 80 dollars" you should have said "the cost to process a claim is approximately normally distributed with a mean of 80 dollars", letting the word "mean" be the only reference to averages.
The null hypothesis would be that the cost-cutting measures had no effect, so that that mean is still at least 80 dollars. The alternative hypothesis were effective, so that the mean is now lower than that.
Generally a null hypothesis is something presumed true until statistical evidence indicates otherwise.
answered Apr 22 '14 at 1:35
Michael Hardy
1
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
I think instead of saying "the average cost to process a claim is approximately normally distributed with a mean of 80 dollars" you should have said "the cost to process a claim is approximately normally distributed with a mean of 80 dollars", letting the word "mean" be the only reference to averages.
The null hypothesis would be that the cost-cutting measures had no effect, so that that mean is still at least 80 dollars. The alternative hypothesis were effective, so that the mean is now lower than that.
Generally a null hypothesis is something presumed true until statistical evidence indicates otherwise.
answered Apr 22 '14 at 1:35
Michael Hardy
1
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
I think instead of saying "the average cost to process a claim is approximately normally distributed with a mean of 80 dollars" you should have said "the cost to process a claim is approximately normally distributed with a mean of 80 dollars", letting the word "mean" be the only reference to averages.
The null hypothesis would be that the cost-cutting measures had no effect, so that that mean is still at least 80 dollars. The alternative hypothesis were effective, so that the mean is now lower than that.
Generally a null hypothesis is something presumed true until statistical evidence indicates otherwise.
answered Apr 22 '14 at 1:35
Michael Hardy
1
I think instead of saying "the average cost to process a claim is approximately normally distributed with a mean of 80 dollars" you should have said "the cost to process a claim is approximately normally distributed with a mean of 80 dollars", letting the word "mean" be the only reference to averages.
The null hypothesis would be that the cost-cutting measures had no effect, so that that mean is still at least 80 dollars. The alternative hypothesis were effective, so that the mean is now lower than that.
Generally a null hypothesis is something presumed true until statistical evidence indicates otherwise.
answered Apr 22 '14 at 1:35
Michael Hardy
1
answered Apr 22 '14 at 1:35
Michael Hardy
1
answered Apr 22 '14 at 1:35
Michael Hardy
1
answered Apr 22 '14 at 1:35
Michael Hardy
1
1
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
– Hani Abdullah
Apr 23 '14 at 16:06
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
Yes. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 16:27
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
How can we test the critical value for this test Sir?
– Hani Abdullah
Apr 23 '14 at 16:29
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
I think that Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance. But i am not sure how to do that. Am i correct in this sir?
– Hani Abdullah
Apr 23 '14 at 16:31
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
You can use $T=dfrac{bar x - 80}{S/sqrt{n}}$ as a test statistic, where $n=25$ is the sample size, $S=10$ is the sample standard deviation, and $bar x=76$ is the sample mean. If the null hypothesis is true this should have a t-distribution with $24$ degrees of freedom. So you would reject the null hypothesis if $T<c$ where $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
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Do you have a question of your own, or are you passing on to us a question written by someone other than yourself by just copying it? Phrasing a question in a manner suitable for assigning homework is frowned upon here.
– Michael Hardy
Apr 21 '14 at 21:55
No sir this not like that. I did solve that question with the help of my teacher but i need to know the exact procedure of it. So could u please help me out?
– Hani Abdullah
Apr 21 '14 at 22:18