Question

You work for a major airline. The airlines overall costs depend very much on price of jet fuel; your task is to estimate the worldwide average price of this fuel. Your airline believes that the price of fuel is 124.9 dollars per barrel and your would like to test this with a hypothesis test. You carry out a simple random sample of n=9 of these prices and find a sample average x(with a line on top of x)= 125.7 dollars per barrel, and find a sample standard deviation of s = 1.034 dollars per barrel. Assume the prices are normally distributed. Test the hypothesis at a = 0.02.

1. Check the assumptions for the hypothesis test.

2. give the null and alternative hypothesis.

3. calculate the test statistic

4. use either p-value or critical value to evaluate

5. conclude the hypothesis test in context

6. construct a 98% confidence interval and interpret briefly

Answer 1

A simple random sample of n=9 of these prices and find a sample average = 125.7 dollars per barrel, and find a sample standard deviation of s = 1.034

r

So we have

n = 9 ; = 125.7   ;    s = 1.034

r

Given level of significance $\alpha$ = 0.02.

Airline believes that the price of fuel is 124.9 dollars per barrel and your would like to test this with a hypothesis test , hence

To Test :-

H0 : $\mu$ = 124.9         { price of fuel may be 124.9 dollars per barrel }

H1 : $\mu$ $\neq$ 124.9      { price of fuel may be significanlty different from 124.9 dollars per barrel }

1. Check the assumptions for the hypothesis test.

1. Since , we have assume the prices are normally distributed , so assumption of normality is satisfied .
2. Here the dependent variable is price of fuel which is a continuous variable , hence assumption the the dependent variable must be continuous is satisfied.
3. We have taken 9 random sample , so each observation is independent to each other , hence assumption of independece in satisfied .

So , the common assumptions for one sample t-test are cheaked .

2. give the null and alternative hypothesis.

To Test :-

H0 : $\mu$ = 124.9         { price of fuel may be 124.9 dollars per barrel }

H1 : $\mu$ $\neq$ 124.9      { price of fuel may be significanlty different from 124.9 dollars per barrel }

3. calculate the test statistic

Test Statistics TS :-

TS =

C 11 s/ n

     = $\frac{125.7-124.9 }{1.034/\sqrt{9}}$

TS = 2.321083

hence caluculated test statistics value is TS = 2.321083

4. use either p-value or critical value to evaluate

To find t-critical value $t_{\alpha /2}$

Here $t_{\alpha /2}$ is t-distributed with n-1 = 9-1 = 8 degree of freedom and $1596227582435_blob.png$ =0.02,

It can be computed from statistical book or more accurately from any software like R,Excel

From R

> qt(1-0.02/2,df=8)
[1] 2.896459

Hence t-critical value is $t_{\alpha /2}$ = 2.896459

Rejection criteria :- We reject null hypothesis if absolute of calculated test statistics is greater than t-critical value $t_{\alpha /2}$

Here | TS | = | 2.321083 | = 2.321083 < 2.896459       ( $t_{\alpha /2}$ = 2.896459 )

i.e | TS | < t-critical value $t_{\alpha /2}$

Using critical value approch to evaluate , we concluded that , since calculated Test statistics value is less than t-critical value , we do not reject null hypothesis

Hence we fail to reject null hypothesis .

5. conclude the hypothesis test in context

Since we do not have enough evidence against null hypothesis H0 i.e we fail to reject H0 , hence we fail to reject the claim of airline believes , and so we conclude that the price of fuel may be 124.9 dollars per barrel .

6. construct a 98% confidence interval and interpret briefly

98% confidence interval is given by

CI = {
r
- $t_{\alpha /2}$ * $\frac{s}{\sqrt{n}}$ ,  
r
+ $t_{\alpha /2}$ * $\frac{s}{\sqrt{n}}$    }

We have n = 9 ; = 125.7   ;    s = 1.034

r

and $1596227582435_blob.png$ =0.02 for 98% confidence , we have obtained t-critical value $t_{\alpha /2}$ = 2.896459

Thus

CI = {
r
- $t_{\alpha /2}$ * $\frac{s}{\sqrt{n}}$ ,  
r
+ $t_{\alpha /2}$ * $\frac{s}{\sqrt{n}}$    }

    = { 125.7 - 2.896459 * $\frac{1.034 }{\sqrt{9}}$ ,   125.7 + 2.896459 * $\frac{1.034 }{\sqrt{9}}$    }

    = { 125.7 - 0.9983129 ,   125.7 + 0.9983129   }

CI = { 124.7017 ,   126.6983   }

Thus 98% confidence interval is { 124.7017 ,   126.6983   }

Interpretation :- By 98% confidence , we can say that we are 98% sure than if this confidence interval does not include the null hypothesis value , then we conclude that there is a statistically significant difference in means i.e we reject null hypothesis .

But , since here 98% confidence interval { 124.7017 , 126.6983   } include the null hypothesis value $\mu$ = 124.9 , hence we fail to reject null hypothesis with 98% confidence .