Question

4. A nutritional scientist is investigating in the difference in the saturated fat content of the double cheeseburgers at two well-known restaurant chains. Rather than rely on the companies’ published data, the researcher collects a random sample of 50 burgers from each of the two restaurant chains for analysis. She finds that those from the first restaurant chain have an average saturated fat content of 12.2 grams per burger, with a standard deviation of 2.6 grams, while those from the second restaurant chain have an average of 11.3 grams per burger, with a standard deviation of 3.1 grams.

(a) Construct a 98% confidence interval for the difference in saturated fat content between the burgers of the two chains.

(b) The second restaurant chain discovers the results of this analysis, and having noticed that the sample mean for their burgers is the lower of the two, wants to launch an advertising campaign on the basis that their double cheeseburgers are healthier (in terms of saturated fat content) than those of their rival. At theα= 0.01 level, test whether this claim is statistically significant.

Answer 1

A nutritional scientist is investigating in the difference in the saturated fat content of the double cheeseburgers at two well-known restaurant chains.

A random sample of 50 burgers from each of the two restaurant chain is taken

Thus n1 = n2 = 50

First restaurant chain have an average saturated fat content of 12.2 grams per burger, with a standard deviation of 2.6 grams, while those from the second restaurant chain have an average of 11.3 grams per burger, with a standard deviation of 3.1 grams.

Thus   1 = 12.2    ,   s1 = 2.6

             2 = 11.3    ,   s2 = 3.1

(a) Construct a 98% confidence interval for the difference in saturated fat content between the burgers of the two chains.

Now confidence interval is given by

CI = { 1 - 2 - *SE , 1 - 2 + *SE }

Now SE =

where sd2 =

Calculation

n1 = n2 = 50

s1 = 2.6

s2 = 3.1

sd2 =

=

sd2 = 8.185

Thus

SE = =

SE = 0.5721888

Now is t-distributed with n1+n2-2 = 98 degree of freedom and =0.02 { for 98% confidence }

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

From R

> qt(1-0.02/2,df=98)
[1] 2.365002

Thus = 2.365002

Now 98% confidence interval will be given by

CI = { 1 - 2 - *SE , 1 - 2 + *SE }

      = { 12.2 - 11.3 - 2.365002 *0.5721888 , 12.2 - 11.3 + 2.365002 *0.5721888 }
      = {   -0.4532277 ,   2.253228 }

So the required 98% confidence interval for the difference in saturated fat content between the burgers of the two chains is {   -0.4532277 ,   2.253228 }

(b) The second restaurant chain discovers the results of this analysis, and having noticed that the sample mean for their burgers is the lower of the two, wants to launch an advertising campaign on the basis that their double cheeseburgers are healthier (in terms of saturated fat content) than those of their rival. At theα= 0.01 level, test whether this claim is statistically significant.

Let u1 and u2 be mean of saturated fat content between the first restaurant and second restaurant respectively .

To Test

H0 : u1 u2    

{ second restaurant chain may not have lower saturated fat content than first resturant }

vs

H1 : u1 > u2    

{ second restaurant chain have lower saturated fat content i.e healthier (in terms of saturated fat content) }

Test Statistics TS :

TS =

Now    1 = 12.2    ,    2 = 11.3   

And    SE = 0.5721888

Hence

TS =

      =

TS = 1.572907

So calculates test statistics is 1.572907

Given level of significance is α= 0.01

Since alternative hyphothesis is of " > " type so this is one tail test ( right tail test )

So t-critical value at α= 0.01 with 98 degree of freedom i.e = is given by

> qt(1-0.01 , df = 98)
[1] 2.365002

Hence t-critical value is 2.365002 .

Now we reject null hypothesis if calculates test statistics value is greater than critical value

Now calculates test statistics is TS = 1.572907 , and = 2.365002

So TS = 1.572907 <   2.365002   ( )

i.e TS <

So we fail to reject null hypothesis at = 0.01 level of significance .

Conclusion : -

We do not reject null hypothesis , so we can say that second restaurant chain may not have lower saturated fat content compare to first resturant .

So the claim of second restaurant chain that they wants to launch an advertising campaign on the basis that their double cheeseburgers are healthier (in terms of saturated fat content) than those of their rival [ first restaurant chain ] , may not be statistical significance i.e their double cheeseburgers may not be more healthier than those of their rival .