A Supermarket Is Interested In Finding Out Whether Th 192212

A Supermarket Is Interested In Finding Out Whether The Mean Weekly Sal

A supermarket is interested in finding out whether the mean weekly sales volume of Coca-Cola are the same when the soft drinks are displayed on the top shelf and when they are displayed on the bottom shelf. 10 stores are randomly selected from the supermarket chain with 5 stores using the top shelf display and 5 stores using the bottom shelf display. Assume that the samples are normally distributed with equal population variances. Refer to the sales volume data in the table below, Top shelf Sales Mean=41.6, Variance=249.84, Bottom shelf sales Mean=62.2, Variance=66.96. What is the t-test statistic?

Paper For Above instruction

The supermarket's inquiry into whether the placement of Coca-Cola on different shelves impacts weekly sales involves a hypothesis test comparing the mean sales of two independent groups: the top shelf and bottom shelf displays. Given the data, the mean sales for the top shelf are 41.6 units with a variance of 249.84, while for the bottom shelf, the mean is 62.2 units with a variance of 66.96. The sample sizes for each group are both 5 stores. The objective is to determine if the observed difference is statistically significant by calculating the t-test statistic for independent samples with equal variances assumed.

To compute the t-test statistic, we start with the formula for independent samples:

t = (x̄₁ - x̄₂) / SE, where SE is the standard error of the difference between means.

Since the variances are assumed equal, the pooled variance (Sp^2) is calculated as:

Sp^2 = [(n₁ - 1) s₁^2 + (n₂ - 1) s₂^2] / (n₁ + n₂ - 2),

where s₁^2 and s₂^2 are the sample variances, and n₁ and n₂ are the sample sizes.

Substituting the provided values:

Sp^2 = [(5 - 1) 249.84 + (5 - 1) 66.96] / (5 + 5 - 2) = (4 249.84 + 4 66.96) / 8 = (999.36 + 267.84) / 8 = 1267.2 / 8 = 158.4.

The standard error (SE) for the difference in means is then:

SE = √[Sp^2 (1/n₁ + 1/n₂)] = √[158.4 (1/5 + 1/5)] = √[158.4 (0.2 + 0.2)] = √[158.4 0.4] = √63.36 ≈ 7.96.

Finally, the t-statistic is:

t = (41.6 - 62.2) / 7.96 = (-20.6) / 7.96 ≈ -2.59.

Thus, the t-test statistic is approximately -2.59, indicating the magnitude and direction of the difference in mean weekly sales between the two shelf placements.

Additional Analysis: Prediction of Twin IQ Based on Regression Line

Apart from the supermarket sales comparison, another statistical analysis involves predicting the IQ of one twin based on the other, using data from 20 twin pairs. The correlation coefficient (r) is 0.865, and the regression equation is y = 3.95 + 0.98x, where x is the IQ of the second twin and y is the IQ of the first twin. The means for the x and y values are 96.87 and 99.05, respectively. Given that the IQ of the second twin is 94, we aim to predict the IQ of the first twin.

Using the regression equation, the predicted IQ (ŷ) of the first twin when the second twin’s IQ (x) is 94 is:

ŷ = 3.95 + 0.98 * 94 = 3.95 + 92.12 = 96.07.

The proximity of this predicted IQ to the mean IQ of 99.05 indicates the model's central tendency, but individual predictions account for variability, and the significance level of 0.05 suggests tests to assess the reliability of this prediction could be performed if necessary.

In conclusion, the calculated t-statistic of approximately -2.59 signifies a statistically significant difference at typical alpha levels (e.g., 0.05), suggesting that shelf placement does influence weekly sales. The regression prediction illustrates how IQ scores of twins are correlated, and the specific IQ of 94 for the second twin leads to a predicted IQ of about 96.07 for the first twin, highlighting the strong positive relationship between twin IQ scores.

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