Question

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\))

Regular Coke and Diet Coke Data Set 26 “Cola Weights and Volumes” in Appendix B includesweights (lb) of the contents of cans of Diet Coke (n= 36,\(\overline x \)= 0.78479 lb, s= 0.00439 lb) and of the contents of cans of regular Coke (n= 36,\(\overline x \)= 0.81682 lb, s= 0.00751 lb).

a. Use a 0.05 significance level to test the claim that the contents of cans of Diet Coke have weights with a mean that is less than the mean for regular Coke.

b. Construct the confidence interval appropriate for the hypothesis test in part (a).

c. Can you explain why cans of Diet Coke would weigh less than cans of regular Coke?

Short answer

a. Reject \({H_0}\), there is a sufficient evidence to support the claim that the contents of cans of Diet Coke have weights with a mean that is less than the mean for regular coke.

b. The 90% confidence interval for the difference of means is between -0.0345 lb and -0.0296 lb.

c. The reason for low weight could be low sugar in the diet coke.

Step-by-step solution

Given information

The samples are independent. The standard deviations are unknown and are not assumed to be equal.

The statistics for the contents of Diet coke and regular coke are stated below:

\({n_1} = 36\),\({\bar x_1} = 0.78479\),\({s_1} = 0.00439\)

\({n_2} = 36\),\({\bar x_2} = 0.81682\), \({s_2} = 0.00751\)

The level of significance is \(\alpha = 0.05\).

State the hypotheses

a. To test the claim that the contents of cans of Diet coke have mean weight less than the mean for regular coke, the null and alternative hypothesis are formulated as:

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

\({H_1}:{\mu _1} < {\mu _2}\)

Here, \({\mu _1},{\mu _2}\) are the population mean contents of weights for diet coke and regular coke respectively.

State the Test statistic

The test statistic \(t\)with degrees of freedom (smaller of \(\left( {{n_1} - 1} \right)\), and \(\left( {{n_2} - 1} \right)\)) is given by,

\(\begin{array}{c}t = \frac{{\left( {\overline {{x_1}} - \overline {{x_2}} } \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\\\ = \frac{{0.78479 - 0.81682}}{{\sqrt {\frac{{{{\left( {0.00439} \right)}^2}}}{{36}} + \frac{{{{\left( {0.00751} \right)}^2}}}{{36}}} }}\\\\ = - 22.092.\end{array}\)

The degree of freedom is computed as,

\(\begin{array}{c}df = \min \left( {\left( {{n_1} - 1} \right),\left( {{n_2} - 1} \right)} \right)\\ = \min \left( {\left( {36 - 1} \right),\left( {36 - 1} \right)} \right)\\ = 35\end{array}\)

State the critical value

The test is left-tailed.

Refer to the standard normal distribution table for the critical value at \(\alpha = 0.05\),and 35 degrees of freedom equal to -1.672.

The decision rule states that if the test statistic is lesser than the critical value, the null hypothesis is rejected.

Here, the test statistic is less than the critical value, which implies that the null hypothesis is rejected at a 0.05 level of significance.

Thus, the null hypothesis is rejected and there is sufficient evidence to support the claim that the mean weight of contents of cans of diet Coke has a mean that is less than that of regular coke.

State the confidence interval estimate

b. The appropriate confidence level corresponding to 0.05 level of significance in the one-tailed test is 90%.

The confidence interval estimate of the difference \({\mu _1} - {\mu _2}\) is:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\)

The critical value is obtained from the t-table with 35 degrees of freedom and 0.10 level of significance as 1.6896. That is,

\(\begin{array}{c}{t_{\frac{\alpha }{2}}} = {t_{\frac{{0.1}}{2}}}\\ = 1.6896\end{array}\)

Substitute the values as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ = 1.6896 \times \sqrt {\frac{{{{\left( {0.00439} \right)}^2}}}{{36}} + \frac{{{{\left( {0.00751} \right)}^2}}}{{36}}} \\ = 0.00245\end{array}\)

Therefore, the 90%confidence interval estimate of the difference \({\mu _1} - {\mu _2}\) is,

\(\begin{array}{c}{\rm{Confidence}}\;{\rm{interval}} = \left( {0.78479 - 0.81682} \right) - 0.00245 < {\mu _1} - {\mu _2} < (0.78479 - 0.81682) + 0.00245\\ = - 0.0345 < {\mu _1} - {\mu _2} < - 0.0296\end{array}\)

Thus, the 90% confidence interval for the difference of means is between -0.0345 lb and -0.0296 lb.

Provide an explanation for low weight of diet coke

c. The confidence interval implies that the weight of diet coke is lesser than that of regular coke as the limits are negative for the confidence interval estimate.

The reason would be low sugar content in the diet coke.