Null Hypothesis
$begingroup$
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 AbdullahHani Abdullah
146
$endgroup$
add a comment |
$begingroup$
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 AbdullahHani Abdullah
146
$endgroup$
2
$begingroup$
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.
$endgroup$
– Michael Hardy
Apr 21 '14 at 21:55
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 21 '14 at 22:18
add a comment |
$begingroup$
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 AbdullahHani Abdullah
146
$endgroup$
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 AbdullahHani Abdullah
146
asked Apr 21 '14 at 21:41
Hani AbdullahHani Abdullah
146
edited Apr 23 '14 at 18:06
Hani Abdullah
asked Apr 21 '14 at 21:41
Hani AbdullahHani Abdullah
146
asked Apr 21 '14 at 21:41
Hani AbdullahHani Abdullah
146
asked Apr 21 '14 at 21:41
Hani AbdullahHani Abdullah
146
146
2
$begingroup$
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.
$endgroup$
– Michael Hardy
Apr 21 '14 at 21:55
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 21 '14 at 22:18
add a comment |
2
$begingroup$
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.
$endgroup$
– Michael Hardy
Apr 21 '14 at 21:55
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 21 '14 at 22:18
2
2
$begingroup$
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.
$endgroup$
– Michael Hardy
Apr 21 '14 at 21:55
$begingroup$
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.
$endgroup$
– Michael Hardy
Apr 21 '14 at 21:55
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 21 '14 at 22:18
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 21 '14 at 22:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 HardyMichael Hardy
1
$endgroup$
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f763706%2fnull-hypothesis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 HardyMichael Hardy
1
$endgroup$
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
$begingroup$
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 HardyMichael Hardy
1
$endgroup$
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
$begingroup$
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 HardyMichael Hardy
1
$endgroup$
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 HardyMichael Hardy
1
answered Apr 22 '14 at 1:35
Michael HardyMichael Hardy
1
answered Apr 22 '14 at 1:35
Michael HardyMichael Hardy
1
answered Apr 22 '14 at 1:35
Michael HardyMichael Hardy
1
1
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
H0: µ = 80 -->null hypothesis... H1: µ < 80 -->alternative hypothesis... So is that correct Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:06
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
Yes. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 16:27
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
How can we test the critical value for this test Sir?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:29
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 23 '14 at 16:31
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
$begingroup$
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 $Pr(T<cmidtext{null})=0.05$. ${}qquad{}$
$endgroup$
– Michael Hardy
Apr 23 '14 at 17:58
|
show 5 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f763706%2fnull-hypothesis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
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.
$endgroup$
– Michael Hardy
Apr 21 '14 at 21:55
$begingroup$
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?
$endgroup$
– Hani Abdullah
Apr 21 '14 at 22:18