Question

Coke and Diet Coke Data Set 26 “Cola Weights and Volumes” in Appendix B includes theweights (in pounds) of cola for a sample of cans of regular Coke (n = 36, \(\bar x\)= 0.81682 lb,s = 0.00751 lb) and the weights of cola for a sample of cans of Diet Coke (n = 36, \(\bar x\) =0.78479 lb, s = 0.00439 lb). Use a 0.05 significance level to test the claim that variation is thesame for both types of Coke.

Short answer

There is sufficient evidence to reject the claim that the population variance of the regular coke group is different from the population variance of the diet coke group.

Step-by-step solution

Given information

For a sample of 36 cans of regular Coke, the mean weight of cola is equal to 0.81682 lb, and the standard deviation is equal to 0.00751 lb. For another sample of 36 cans of Diet Coke, the mean weight of cola is equal to 0.78479 lb,and the standard deviation is equal to 0.00439 lb.

It is claimed that the variation in the amount of cola for the regular Coke is equal to the variation in the amount of Cola for the Diet Coke.

Hypotheses

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviations of the weight of colafor the regular Coke and the Diet Cokerespectively.

Null Hypothesis:The populationstandard deviationof the weight of colafor the regular Cokeis equal to the population standard deviation of the weight of colafor the Diet Coke.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternative Hypothesis:The populationstandard deviationof the weight of colafor the regular Cokeis not equal to the population standard deviationof the weight of colafor the Diet Coke.

Symbolically,

\({H_1}:{\sigma _1} \ne {\sigma _2}\)

Compute the test statistic

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {0.00751} \right)^2} = 0.000056\)

\({\left( {0.00439} \right)^2} = 0.000019\)

Here,\(s_1^2\)is the sample variance corresponding to the regular Coke has a value equal to 0.000056 lb squared.

\(s_2^2\)is the sample variance corresponding to the Diet Cokeand has a value equal to 0.000019 lb squared.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{{{\left( {0.00751} \right)}^2}}}{{{{\left( {0.00439} \right)}^2}}}\\ = 2.927\end{array}\)

Thus, F is equal to 2.927.

Critical value and p-value

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 36 - 1\\ = 35\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 36 - 1\\ = 35\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 35 and denominator degrees of freedom equal to 35 for a right-tailed test.

The level of significance is equal to:

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Thus, the critical value is equal to 1.9611.

The two-tailed p-value for F equal to 2.927 is equal to 0.0020.

Conclusion

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to rejectthe claimthat variation is the same for both types of Coke.