Question

An advertisement for a popular supermarket chain claims that it has had consistently lower prices than one of its competitors. As part of a survey conducted by an independent price-checking company, the average weekly total, based on the prices of approximately 95 items, is given for this chain and for its competitor recorded during four consecutive weeks in a particular month. $$ \begin{array}{lcc} \text { Week } & \text { Advertiser (\$) } & \text { Competitor (\$) } \\ \hline 1 & 254.26 & 256.03 \\ 2 & 240.62 & 255.65 \\ 3 & 231.90 & 255.12 \\ 4 & 234.13 & 261.18 \end{array} $$ a. Is there a significant difference in the average prices for these two different supermarket chains? b. What is the approximate \(p\) -value for the test conducted in part a? c. Construct a \(99 \%\) confidence interval for the difference in the average prices for the two supermarket chains. Interpret this interval.

Short answer

Answer: Yes, there is a significant difference in the average prices of the two supermarket chains. The p-value is very small (less than 0.001), providing strong evidence against the null hypothesis. Additionally, the 99% confidence interval for the difference in average prices, approximately (-30.2525, -3.2825), suggests that the Advertiser has significantly lower prices than the Competitor.

Step-by-step solution

Formulate the null and alternative hypotheses

The null hypothesis (H0) states that there is no significant difference in the average prices for the two supermarket chains, while the alternative hypothesis (H1) states that there is a significant difference in the average prices for the two supermarket chains. $$ H_0: \mu_1 - \mu_2 = 0 \\ H_1: \mu_1 - \mu_2 \neq 0 $$ Where \(\mu_1\) denotes the average price for the Advertiser and \(\mu_2\) denotes the average price for the Competitor.

Calculate the mean and standard deviations

Calculate the means and standard deviations for both supermarket chains using the given data. Mean (`Advertiser`) = \(\frac{254.26+240.62+231.9+234.13}{4} = 240.2275\) Mean (`Competitor`) = \(\frac{256.03+255.65+255.12+261.18}{4} = 256.995\) Standard Deviation (`Advertiser`) = \(s_1 = \sqrt{\frac{(254.26-240.2275)^2+(240.62-240.2275)^2+(231.9-240.2275)^2+(234.13-240.2275)^2}{4-1}} = 10.034\) Standard Deviation (`Competitor`) = \(s_2 = \sqrt{\frac{(256.03-256.995)^2+(255.65-256.995)^2+(255.12-256.995)^2+(261.18-256.995)^2}{4-1}} = 2.713\)

Conduct the hypothesis test

Using the t-test for independent samples, compute the test statistic value. \(t = \frac{(\bar{X_1} - \bar{X_2}) - (\mu_1 - \mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}} = \frac{(240.2275 - 256.995) - 0}{\sqrt{\frac{10.034^2}{4}+\frac{2.713^2}{4}}} = -5.919\) The degrees of freedom for the t-test can be estimated using the following equation: \(df \approx \frac{(\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2})^2}{\frac{(\frac{s^2_1}{n_1})^2}{(n_1 - 1)} + \frac{(\frac{s^2_2}{n_2})^2}{(n_2 - 1)}} = \frac{(\frac{10.034^2}{4} + \frac{2.713^2}{4})^2}{\frac{(\frac{10.034^2}{4})^2}{3} + \frac{(\frac{2.713^2}{4})^2}{3}} = 5.61 \approx 6\) The test statistic value is \(-5.919\) and the degrees of freedom are approximately \(6\).

Find the p-value

Using a t-distribution table or an online calculator, find the p-value corresponding to the test statistic value \(t = -5.919\) and \(df = 6\). The p-value should be very small (less than 0.001). This indicates strong evidence against the null hypothesis.

Compute the 99% confidence interval

Using the t-distribution, calculate the 99% confidence interval for the difference in average prices. CI = \((\bar{X_1} - \bar{X_2}) \pm t_{\alpha/2, df} \times \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}\) CI = \((240.2275 - 256.995) \pm 3.707 \times \sqrt{\frac{10.034^2}{4}+\frac{2.713^2}{4}}\) = \((-16.7675 \pm 13.485)\). The 99% confidence interval for the difference in average prices is approximately \((-30.2525, -3.2825)\). This interval suggests that the Advertiser has prices significantly lower than the Competitor.

Key concepts

Confidence Interval

A confidence interval provides a range of values that is likely to contain the population parameter we are interested in. When we say "99% confidence interval," it means that if we conducted the same study 100 times, in approximately 99 of them, the calculated interval would include the true difference in average prices between the two supermarkets.

In our problem, the 99% confidence interval for the price difference is from \(-30.2525\) to \(-3.2825\). This interval indicates that, with high confidence, the average price of items at the advertiser's store is lower than that of the competitor. Essentially, if the entire interval is negative, it suggests a consistent price advantage for the advertiser.

Understanding this concept is like being given a fishing net (the interval) and catching the same kind of fish (the parameter) with near certainty each time you throw it in the lake (conduct the study). By capturing a range, rather than a single point, we acknowledge the role of variability and chance in any statistical estimate.

T-Distribution

The t-distribution is a fundamental concept in statistics, used particularly when dealing with small sample sizes. It resembles the normal distribution but has thicker tails, meaning it's more spread out. This accounts for the greater uncertainty we face when estimating with smaller samples.

In hypothesis testing, especially involving means, we often assume distribution of the test statistics follows a t-distribution. This is precisely why we use the t-distribution tables or tools to find critical values and p-values when our sample size is not large.

In the solution above, we use the t-distribution to determine the test statistic and to calculate the confidence interval. Its flexibility with smaller samples makes it indispensable for cases with limited data, as it adjusts for the additional variability expected in these situations.

P-value

The p-value helps in decision-making in hypothesis tests. It measures the strength of evidence against the null hypothesis. A small p-value, typically less than 0.05, indicates strong evidence against the null hypothesis and suggests that you should reject it.

For our case, the calculated p-value is less than 0.001. This value signals overwhelming evidence that the average prices of the two supermarkets are indeed different, prompting rejection of the null hypothesis.

Imagine the p-value as a measure of surprise. If the null hypothesis were true, and you ended up with data resulting in such a low p-value, you'd be shocked. This shock leads us to believe the null might not be true, echoing the idea that "this can't be just by chance."

Independent Samples

When conducting a statistical test, the term "independent samples" refers to data from two separate groups where the measurements in one group do not affect those in the other. This is essential for certain types of hypothesis testing, as seen in this exercise.

In our supermarket price comparison, the weekly prices collected for the advertiser and its competitor are considered independent of each other. This independence is crucial for using the two-sample t-test, as it ensures unbiased results without confounding influences.

Think of independent samples as two unconnected puzzle sets. Trying to solve them together won't work as they don’t fit with each other; they are individual sets with their own pieces and picture to complete.

T-Test

A t-test is a statistical test used to determine if there is a significant difference between the means of two groups, which may be related in certain features. In the exercise at hand, a two-sample t-test helps to determine if the average prices of the advertiser and its competitor vary significantly.

The data includes comparing the average prices across two groups to see if the difference found is meaningful or could just happen by chance. With our t-test, this difference was tested and found to be significant, suggesting the advertiser's prices are truly lower.

The t-test essentially asks "are the means of these two groups different enough," considering the spread of the data and the number of samples, to rule out random variation as the cause? In our exercise, the answer was yes, leading to conclusions about pricing strategies between these competitors.