91影视

The given problem is based on the data of Coke and Pepsi. This data set contains information about the volume of the contents in cans of regular Coke and regular Pepsi, summarized as follows:

$$\begin{gathered} {\text{1}}^{\text{st}}\text{sample}\text{:Regular}\text{coke} \\ {n}_1=36, \\ {s}_1=0.11oz \\ {\stackrel{炉}{x}}_1=12.19oz \end{gathered}$$

$$\begin{gathered} {\text{2}}^{\text{nd}}\text{sample}:\text{Regular}\text{Pepsi} \\ {n}_2=20 \\ {s}_2=0.09oz \\ {\stackrel{炉}{x}}_2=12.29oz \end{gathered}$$

a.

The claim is states that the mean volume of a can of Coke and a can of Pepsi is the same.

$$\begin{gathered} H_0:{渭}_1={渭}_2 \\ {H}_1:{渭}_1鈮�{渭}_2 \end{gathered}$$

Here,${渭}_1,{渭}_2$are the population mean volume contents for Coke and Pepsi, respectively.

The samples are independent with unknown and unequal population standard deviations.

The formula for t-statistic is given below.

$t_{stat}=\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}};(\text{here},\left({渭}_1-{渭}_2\right)\text{is}\text{supposed}\text{to}\text{be}0鈥�)$

For t-distribution, find the degrees of freedom as follows:

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

The critical values are obtained as follows:

$$\begin{gathered} P\left(t>t_{\frac{伪}{2}}\right)=\frac{伪}{2} \\ P\left(t>t_{\frac{0.05}{2}}\right)=\frac{0.05}{2} \\ P\left(t>t_{0.025}\right)=0.025 \end{gathered}$$

Thus, the critical value obtained from the t-table for 35 degrees of freedom is 2.0301.

The test statistic of the means of populations is as follows:

$$\begin{gathered} t_{stat}=\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{\left(12.19-12.29\right)}{\sqrt{\frac{{\left(0.11\right)}^2}{36}+\frac{{\left(0.09\right)}^2}{36}}} \\ =-4.22159 \end{gathered}$$

The test statistic is $t_{stat}=-4.22$.

The decision criterion for this problem statement is given below.

If $\left|t_{stat}\right|>t_{crit}$; Reject a null hypothesis at level of significance

If $\left|t_{stat}\right|<t_{crit}$; Fail to accept null hypothesis at level of significance

In this case, $\left|t_{stat}=-4.222\right|>t_{crit}=2.0301$.

Thus, the null hypothesis is rejected. It shows that there is not enough evidence to support the claim that the volumes of cans of Pepsi and Coke are the same.

b.

For the 0.05 significance test, concerning a two-tailed test, the most appropriate confidence level is 95%.

The formula for the confidence interval of the means of population is given by

$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-E<\left({渭}_1-{渭}_2\right)<\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)+E$

Here, E is the margin of error that is computed as follows:

$$\begin{gathered} E=t_{\frac{伪}{2}}脳\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \\ =t_{\frac{0.05}{2}}脳\sqrt{\frac{{0.11}^2}{36}+\frac{{0.09}^2}{36}} \\ =2.0301脳0.024 \\ =0.0481 \end{gathered}$$

Substitute the values in the formula.

$$\begin{gathered} \text{C.I}=\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-E<\left({渭}_1-{渭}_2\right)<\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)+E \\ =\left(12.19-12.29\right)-0.59<\left({渭}_1-{渭}_2\right)<\left(12.19-12.29\right)+0.59 \\ =-0.052<\left({渭}_1-{渭}_2\right)<0.148 \end{gathered}$$

The 95% confidence interval includes 0, which implies that there is not enough evidence that the mean of the contents for Pepsi and Coke is the same.

There is a significant difference between the volumes of the cans.

c.

Thus, it is concluded that there is a statistical significance for the difference in mean of the contents in Pepsi and Coke.

There does not appear to be any practical significance as the difference in the sample means is 0.10 oz, which is very small.