Questions: A student compared organic food prices at Grocery 1 and Grocery 2. The same items were priced at each store. The first three items are shown in the accompanying Figure A. Choose the correct accompanying output ( B or C ) for the appropriate test, explaining why you chose that output. Then test the hypothesis that the population means are not equal using a significance level of 0.05. Assume that all necessary conditions for carrying out a hypothesis test hold. Figure A and Outputs (Figure A) Food Grocery 1 Grocery 2 Bananas/1 lb 0.89 0.89 Grape tomatoes/1 lb 4.39 4.09 Russet potato/5 lb 4.49 4.89 (Figure B) Paired T-Test and CI: Grocery N Mean StDev SE Mean Grocery 1 39 3.01 1.34 0.21 Grocery 2 39 3.13 1.38 0.22 Difference 39 -0.12 1.32 0.21 95% CI for difference: (-0.548, 0.308) T-Test of mean difference = 0 (vs ≠): T-Value = -0.57 P-value = 0.574 (Figure C) Two-sample T-Test and CI: Grocery 1, Grocery 2 N Mean StDev SE Mean Grocery 1 39 3.01 1.34 0.21 Grocery 2 39 3.13 1.38 0.22

A student compared organic food prices at Grocery 1 and Grocery 2. The same items were priced at each store. The first three items are shown in the accompanying Figure A. Choose the correct accompanying output ( B or C ) for the appropriate test, explaining why you chose that output. Then test the hypothesis that the population means are not equal using a significance level of 0.05. Assume that all necessary conditions for carrying out a hypothesis test hold.

Figure A and Outputs
(Figure A)
Food  Grocery 1  Grocery 2
Bananas/1 lb  0.89  0.89
Grape tomatoes/1 lb  4.39  4.09
Russet potato/5 lb  4.49  4.89

(Figure B)
Paired T-Test and CI: Grocery
 N  Mean  StDev  SE Mean
Grocery 1  39  3.01  1.34  0.21
Grocery 2  39  3.13  1.38  0.22
Difference  39  -0.12  1.32  0.21

95% CI for difference: (-0.548, 0.308)
T-Test of mean difference = 0 (vs ≠): T-Value = -0.57  P-value = 0.574

(Figure C)
Two-sample T-Test and CI: Grocery 1, Grocery 2
 N  Mean  StDev  SE Mean
Grocery 1  39  3.01  1.34  0.21
Grocery 2  39  3.13  1.38  0.22
Transcript text: A student compared organic food prices at Grocery 1 and Grocery 2. The same items were priced at each store. The first three items are shown in the accompanying Figure A. Choose the correct accompanying output ( B or C ) for the appropriate test, explaining why you chose that output. Then test the hypothesis that the population means are not equal using a significance level of 0.05. Assume that all necessary conditions for carrying out a hypothesis test hold. Figure A and Outputs \begin{tabular}{|lll|} \hline (Figure A) & & \\ \hline Food & Grocery 1 & Grocery 2 \\ Bananas/1 lb & 0.89 & 0.89 \\ Grape tomatoes/1 lb & 4.39 & 4.09 \\ Russet potato/5 lb & 4.49 & 4.89 \\ \hline \end{tabular} (Figure B) \begin{tabular}{|lrrrr|} \multicolumn{5}{c|}{ Paired T-Test and Cl: Grocery } \\ \hline & N & Mean & StDev & SE Mean \\ Grocery 1 & 39 & 3.01 & 1.34 & 0.21 \\ Grocery 2 & 39 & 3.13 & 1.38 & 0.22 \\ Difference & 39 & -0.12 & 1.32 & 0.21 \\ \hline \end{tabular} 95\% Cl for difference: ( $-0.548,0.308$ ) T -Test of mean difference $=0($ vs $\neq): T$-Value $=-0.57 \quad$ P-value $=0.574$ (Figure C) Two-sample T-Test and Cl: Grocery 1, Grocery 2 \begin{tabular}{|lrrrr|} \hline & N & Mean & StDev & SE Mean \\ Grocery 1 & 39 & 3.01 & 1.34 & 0.21 \\ Grocery 2 & 39 & 3.13 & 1.38 & 0.22 \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Define the Hypothesis

We are testing the hypothesis that the population means of organic food prices at Grocery 1 and Grocery 2 are equal. The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)) are defined as follows:

\[ H_0: \mu_1 - \mu_2 = 0 \quad \text{(the means are equal)} \] \[ H_a: \mu_1 - \mu_2 \neq 0 \quad \text{(the means are not equal)} \]

Step 2: Calculate the Test Statistic

The mean difference (\(D\)) between the paired samples is calculated as:

\[ D = \bar{x}_1 - \bar{x}_2 = -0.0333 \]

The standard deviation of the differences is:

\[ s_D = 0.3512 \]

The standard error (SE) of the mean difference is calculated using:

\[ SE = \frac{s_D}{\sqrt{n}} = \frac{0.3512}{\sqrt{3}} = 0.2028 \]

The test statistic (\(t\)) is then calculated as:

\[ t = \frac{D}{SE} = \frac{-0.0333}{0.2028} = -0.1644 \]

Step 3: Determine the Critical Value and P-Value

For a two-tailed test at a significance level of \(\alpha = 0.05\) with \(df = n - 1 = 2\), the critical value is:

\[ t_{\alpha/2, df} = t_{(0.025, 2)} = 4.3027 \]

The p-value is calculated as:

\[ P = 2 \times (1 - T(|t|)) = 2 \times (1 - T(0.1644)) = 0.8845 \]

Step 4: Make a Decision

We compare the p-value to the significance level:

\[ P = 0.8845 > \alpha = 0.05 \]

Since the p-value is greater than the significance level, we fail to reject the null hypothesis.

Final Answer

There is not enough evidence to say that the population means are not equal. Thus, the conclusion is:

\[ \boxed{H_0 \text{ is not rejected}} \]

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