/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 An advertisement for Albertsons,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An advertisement for Albertsons, a supermarket chain in the western United States, claims that Albertsons has had consistently lower prices than four other fullservice supermarkets. As part of a survey conducted by an "independent market basket price-checking company," the average weekly total, based on the prices of approximately 95 items, is given for two different supermarket chains recorded during 4 consecutive weeks in a particular month. Week Albertsons Ralphs \(\begin{array}{lll}1 & 254.26 & 256.03 \\ 2 & 240.62 & 255.65 \\ 3 & 231.90 & 255.12\end{array}\) \(\begin{array}{lll}4 & 234.13 & 261.18\end{array}\) 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

Expert verified
Answer: Yes, there is a significant difference in the average prices of Albertsons and Ralphs supermarket chains. The 99% confidence interval (-26.573, -16.962) indicates that Albertsons' average prices are lower than Ralphs' average prices by an amount between 16.962 and 26.573.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (H0) states that there is no significant difference in the average prices of the two supermarket chains, while the alternative hypothesis (H1) claims that there is a significant difference in the average prices. Mathematically: H0: \(\mu_A = \mu_R\) H1: \(\mu_A \neq \mu_R\) where \(\mu_A\) is the average price for Albertsons and \(\mu_R\) is the average price for Ralphs.
02

Calculate the test statistic

To conduct the t-test, we first need to calculate the mean and standard deviation for each supermarket chain and then compute the test statistic. Mean of Albertsons (\(\bar{x}_A\)): \(\frac{254.26 + 240.62 + 231.90 + 234.13}{4} = \frac{960.91}{4} = 240.2275\) Mean of Ralphs (\(\bar{x}_R\)): \(\frac{256.03 + 255.65 + 255.12 + 261.18}{4} = \frac{1027.98}{4} = 256.995\) Standard deviation of Albertsons (\(s_A\)): \(\sqrt{\frac{(240.2275-254.26)^2+(240.2275-240.62)^2+(240.2275-231.9)^2+(240.2275-234.13)^2}{4-1}} = 9.691\) Standard deviation of Ralphs (\(s_R\)): \(\sqrt{\frac{(256.995-256.03)^2+(256.995-255.65)^2+(256.995-255.12)^2+(256.995-261.18)^2}{4-1}} = 3.102\) Using the formula for the t-test statistic: \(t = \frac{(\bar{x}_A - \bar{x}_R)}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_R^2}{n_R}}}\) where \(n_A\) and \(n_R\) are the sample sizes for Albertsons and Ralphs, respectively.
03

Compute the test statistic

Plugging in the values, we have: \(t = \frac{240.2275 - 256.995}{\sqrt{\frac{9.691^2}{4} + \frac{3.102^2}{4}}} = -7.794\) #b. Calculating the approximate p-value#
04

Calculate the degrees of freedom

In order to find the p-value, we need to determine the degrees of freedom (df). For the two-sample t-test, the degrees of freedom are given by: \(df = n_A + n_R - 2 = 4 + 4 - 2 = 6\)
05

Approximate the p-value

Using a t-table or t-distribution calculator and inputting our t-test statistic and degrees of freedom, we find that the p-value is approximately 0.0005, which is very small. #c. Constructing a 99% confidence interval#
06

Find the t-critical value

To construct the 99% confidence interval, we need to find the t-critical value, which corresponds to a 99% level of confidence and 6 degrees of freedom. Checking a t-table or using a calculator, we find the t-critical value is approximately 3.707.
07

Calculate the confidence interval

The confidence interval is given by: \(\bar{x}_A - \bar{x}_R \pm t_{critical} \cdot{\sqrt{\frac{s_A^2}{n_A} + \frac{s_R^2}{n_R}}}\) Plugging in the values, we have: \(240.2275 - 256.995 \pm 3.707 \cdot{\sqrt{\frac{9.691^2}{4} + \frac{3.102^2}{4}}}\) The resulting interval is approximately \((-26.573, -16.962)\).
08

Interpret the confidence interval

Since the 99% confidence interval for the difference in the average prices does not contain 0, we can conclude that there is a significant difference between the average prices of these two supermarket chains. The interval (-26.573, -16.962) indicates that, with 99% confidence, Albertsons' average prices are lower than Ralphs' average prices by an amount between \(16.962 and \)26.573.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is a powerful statistical tool used to make decisions or inferences about a population based on a sample. This process begins with the formation of two competing hypotheses: the null hypothesis ( H0 ), which suggests no effect or no difference, and the alternative hypothesis ( H1 ), which indicates the presence of an effect or a difference.

In this scenario involving Albertsons and Ralphs, the null hypothesis claims there is no significant difference in average prices between the two supermarket chains. The alternative hypothesis suggests otherwise, proposing that a significant difference does exist. To perform hypothesis testing, we calculate a test statistic and compare it with a threshold (or critical value).

The results tell us whether we "reject" the null hypothesis or "fail to reject" it. This decision is usually made by looking at the p-value—a smaller p-value (typically less than 0.05) implies that the data contradicts the null hypothesis, leading to its rejection. Hypothesis testing helps objectify decisions using statistical evidence rather than mere speculation.
Confidence Interval
A confidence interval provides an estimated range that is believed to contain the true difference between two population means with a certain level of confidence. In our example, we are asked to construct a 99% confidence interval around the average price difference between Albertsons and Ralphs.
To calculate this interval, we use the formula:\[ \bar{x}_A - \bar{x}_R \pm t_{critical} \cdot{\sqrt{\frac{s_A^2}{n_A} + \frac{s_R^2}{n_R}}} \]This formula considers the observed difference in means, the variability of the data (standard deviations), and the size of the samples (number of weeks observed).
Our confidence interval gives us a range—for instance, (-26.573, -16.962)—which means that we can be 99% confident that the actual average price difference between the supermarkets falls within this interval.
Where confidence intervals are concerned, if the interval does not include zero, it suggests a significant difference, aligning with the conclusions drawn from hypothesis testing.
Degrees of Freedom
Degrees of freedom are a concept that helps us to understand the number of independent values in a statistical calculation that can vary. When conducting a t-test, degrees of freedom are an essential factor as they influence the shape of the t-distribution.
In the context of comparing the average prices of Albertsons and Ralphs, the degrees of freedom are calculated based on the number of observations from each group, using the formula:\[ df = n_A + n_R - 2 \]For our example, with four observations from each supermarket, the degrees of freedom equate to 6.
Degrees of freedom are crucial in determining the critical t-value for hypothesis testing and the construction of confidence intervals. This value is sourced from a t-distribution table, tailored to the specified degrees of freedom, to ensure that the statistical testing accurately reflects the variability in the data.
P-value
The p-value is a measure that helps us determine the strength of the evidence against the null hypothesis. It is the probability of obtaining test results at least as extreme as those observed, under the assumption that the null hypothesis is true.

In the supermarket price comparison, the calculated p-value was approximately 0.0005. This very low p-value indicates a strong evidence against the null hypothesis, suggesting that the average prices are indeed significantly different between Albertsons and Ralphs.
Common thresholds for p-values, such as 0.05 or 0.01, help determine when to reject the null hypothesis. A p-value lower than these thresholds indicates that the observed data would be highly unlikely if the null hypothesis were correct, thus reinforcing the decision to reject it.
  • A small p-value (e.g., < 0.05) typically indicates strong evidence against the null hypothesis, leading to its rejection.
  • A large p-value suggests weak evidence against the null hypothesis, so it is not rejected.
Understanding p-values is crucial, as they bridge the gap between anecdotal observations and scientifically valid conclusions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the Student's \(t\) Probabilities applet to find the following critical values: a. an upper one-tailed rejection region with \(\alpha=.05\) and 11 df b. a two-tailed rejection region with \(\alpha=.05\) and 7 df c. a lower one-tailed rejection region with \(\alpha=.01\) and \(15 d f\)

In a study of the infestation of the Thenus orientalis lobster by two types of barnacles, Octolasmis tridens and \(O .\) lowei, the carapace lengths (in millimeters) of 10 randomly selected lobsters caught in the seas near Singapore are measured: \(\begin{array}{llll}78 & 66 & 65 & 63\end{array}\) \(\begin{array}{lll}60 & 60 & 58\end{array}\) $$ \begin{array}{lll} 56 & 52 & 50 \end{array} $$ Find a \(95 \%\) confidence interval for the mean carapace length of the \(T\). orientalis lobsters.

An experiment published in The American Biology Teacher studied the efficacy of using \(95 \%\) ethanol or \(20 \%\) bleach as a disinfectant in removing bacterial and fungal contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using eggplant as the plant tissue being cultured. \({ }^{8}\) Five cuttings per plant were placed on a petridish for each disinfectant and stored at \(25^{\circ} \mathrm{C}\) for 4 weeks. The observation reported was the number of uncontaminated eggplant cuttings after the 4 -week storage. Disinfectant \(95 \%\) Ethanol \(20 \%\) Bleach \begin{tabular}{lcc} \hline Mean & 3.73 & 4.80 \\ Variance & 2.78095 & .17143 \\ \(n\) & 15 & 15 \\ & Pooled variance 1.47619 & \end{tabular} a. Are you willing to assume that the underlying variances are equal? b. Using the information from part a, are you willing to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested?

At a time when energy conservation is so important, some scientists think closer scrutiny should be given to the cost (in energy) of producing various forms of food. Suppose you wish to compare the mean amount of oil required to produce 1 acre of corn versus 1 acre of cauliflower. The readings (in barrels of oil per acre), based on 20-acre plots, seven for each crop, are shown in the table. Use these data to find a \(90 \%\) confidence interval for the difference between the mean amounts of oil required to produce these two crops. Corn Cauliflower \(\begin{array}{ll}5.6 & 15.9\end{array}\) 7.1 13.4 \(\begin{array}{ll}4.5 & 17.6\end{array}\) 6.0 16.8 7.9 15.8 \(\begin{array}{ll}4.8 & 16.3\end{array}\) 57 \(17 .\)

Two methods were used to measure the specific activity (in units of enzyme activity per milligram of protein) of an enzyme. One unit of enzyme activity is the amount that catalyzes the formation of 1 micromole of product per minute under specified conditions. Use an appropriate test or estimation procedure to compare the two methods of measurement. Comment on the validity of any assumptions you need to make. $$ \begin{array}{l|lllll} \text { Method } 1 & 125 & 137 & 130 & 151 & 142 \\\ \hline \text { Method } 2 & 137 & 143 & 151 & 156 & 149 \end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.