Problem 50 The authors of the paper "Change... [FREE SOLUTION]

Chapter 12: Problem 50

The authors of the paper "Changesin Quantity, Spending, and Nutritional Characteristics of Adult, Adolescent and Child Urban Corner Store Purchases After an Environmental Intervention"(Preventative Medicine [2015]: $81-85$ ) wondered if increasing the availability of healthy food options would also increase the amount people spend at the corner store. They collected data from a representative sample of 5949 purchases at corner stores in Philadelphia after the stores increased their healthy food options. The sample mean amount spent for this sample of purchases was $\$ 2.86$ and the sample standard deviation was $\$ 5.40$. a. Notice that for this sample, the sample standard deviation is greater than the sample mean. What does this tell you about the distribution of purchase amounts? b. Before the stores increased the availability of healthy foods, the population mean total amount spent per purchase was thought to be about $\$ 2.80$. Do the data from this study provide convincing evidence that the population mean amount spent per purchase is greater after the change to increased healthy food options? Carry out a hypothesis test with a significance level of 0.05

Short Answer

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The distribution of purchase amounts is highly dispersed and potentially skewed, with a sample mean of $2.86 and a sample standard deviation of $5.40. Based on the hypothesis test with a significance level of 0.05, we have enough evidence to reject the null hypothesis and conclude that the population mean amount spent per purchase is greater after increasing the availability of healthy food options at corner stores.

Step by step solution

a. Interpretation of sample mean and standard deviation

The sample mean represents the average amount spent per purchase in the sample, which is $2.86. The sample standard deviation measures the dispersion or spread of the data. A standard deviation of $5.40 indicates that the purchase amounts tend to be more spread out from the mean. In this context, having a sample standard deviation greater than the sample mean reveals that the distribution of purchase amounts is highly dispersed and potentially skewed, where many small purchases and a few high-priced purchases could lead to a high standard deviation and skew the distribution to the right.

b. Hypothesis Test

We are testing if the sample provides convincing evidence that there is an increase in the population mean amount spent per purchase after increasing the availability of healthy food options. To perform the hypothesis test, we will follow these steps: Step 1: Set the hypotheses: The null hypothesis (H$_0$): $\mu = \$ 2.80$ The alternative hypothesis (H$_1$): $\mu > \$ 2.80$ Step 2: Choose the significance level: $\alpha = 0.05$ Step 3: Calculate the test statistic: The test statistic for a one-sample t-test is: \[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\] where $\bar{x}$ is the sample mean, $\mu_0$ is the population mean under H$_0$, $s$ is the sample standard deviation, and $n$ is the sample size. In our case, we have: $n = 5949$ $\bar{x} = \$ 2.86$ $\mu_0 = \$ 2.80$ $s = \$ 5.40$ So the test statistic is: \[t = \frac{2.86 - 2.80}{\frac{5.40}{\sqrt{5949}}}\] Calculating the value, we get: \[t \approx 2.52 \] Step 4: Determine the critical value and make a decision: Since we are conducting a one-tailed test with $\alpha = 0.05$, we need to determine the critical value from the t-distribution table. With a large sample size, we can use the z-table instead. For a one-tailed test at $\alpha = 0.05$, the critical value is 1.645. Our test statistic, $t \approx 2.52 > 1.645$, is greater than the critical value, so we reject the null hypothesis. Step 5: Conclude Based on the hypothesis test with a significance level of 0.05, we have enough evidence to reject the null hypothesis and conclude that the population mean amount spent per purchase is greater after increasing the availability of healthy food options at corner stores.

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Key Concepts

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

Sample Mean

The sample mean is an important concept in statistics, especially when analyzing data from samples. In this case, the sample mean represents the average amount of money spent per purchase at corner stores, which is calculated to be $2.86. This is found by adding up the total amount spent on all purchases in the sample and then dividing by the number of purchases (5949 in this exercise).

This value gives us a centralized measure of spending within the sample and serves as a key figure when comparing the sample to the wider population. The sample mean is used to estimate the population mean, providing insights into consumer behavior post-intervention in corner stores.

Sample Standard Deviation

The sample standard deviation is a measure of how much the individual purchase amounts deviate from the sample mean. In this case, the standard deviation is $5.40, which is notably higher than the sample mean. This suggests a wide spread in purchase amounts.

When the standard deviation is greater than the mean, it indicates high variability in the data. This might point to a right-skewed distribution in the context of this exercise. A possible scenario could include frequent smaller purchases, but with occasional larger ones that elevate the standard deviation. This understanding of data dispersion helps set the stage for further statistical analysis.

Significance Level

The significance level is a crucial part of hypothesis testing. It represents the threshold probability for rejecting the null hypothesis. In this exercise, a significance level ($\alpha$) of 0.05 is chosen. This means there's a 5% risk of concluding that the population mean has increased when, in fact, it has not.

A significance level of 0.05 is a common choice in many statistical studies, balancing the risk of errors in hypothesis testing. It helps establish a critical value that the test statistic must exceed to reject the null hypothesis. This level of significance ensures that the conclusions drawn are reasonably robust and persuasive.

T-Test

The t-test is a statistical method used to determine if there is a significant difference between the sample mean and a known population mean. In this exercise, the focus is on whether the mean amount spent per purchase has increased post-intervention.

We follow several crucial steps for the t-test:

Formulate the null hypothesis ($ H_0: \mu = \\(2.80 $) and the alternative hypothesis ($ H_1: \mu > \$2.80 \)).
Set the significance level at 0.05.
Calculate the test statistic using the formula: \[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \]
Compare the test statistic to the critical value.

With a calculated t-statistic of approximately 2.52, and a critical value of 1.645, the test statistic exceeds the critical boundary, leading to a rejection of the null hypothesis. Thus, the evidence strongly suggests an increase in the population mean amount spent.

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Short Answer

Step by step solution

a. Interpretation of sample mean and standard deviation

b. Hypothesis Test

Key Concepts

Sample Mean

Sample Standard Deviation

Significance Level

T-Test

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