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Chapter 9: Problem 61

According to a study conducted in \(2015,18 \%\) of shoppers said that they prefer to buy generic instead of name-brand products. Suppose that in a recent sample of 1500 shoppers, 315 stated that they prefer to buy generic instead of name-brand products. At a \(5 \%\) significance level, can you conclude that the proportion of all shoppers who currently prefer to buy generic instead of name-brand products is higher than . 18 ? Use both the \(p\) -value and the critical-value approaches.

Short Answer

Expert verified
Conclusion will be based on the comparison of calculated p-value and significance level.

Step by step solution

01

Null and Alternate Hypothesis

In this case, the null hypothesis will be that the prevalence of preferring generic products to brand names is still 18% (\(H_0: p = 0.18\)). The alternate hypothesis is that the prevalence has increased and it is more than 18% (\(H_a: p > 0.18\)). Here, \(p\) is the proportion of all shoppers who prefer to buy generic instead of name-brand products.
02

Test Statistic Calculation

Let's calculate the test statistic 'z' using the formula for testing a population proportion: \(z = (p' - p_0) / \sqrt{p_0 * (1 - p_0) / n}\), where \(p'\) is the sample proportion, \(p_0\) is the proportion from the null hypothesis, and \(n\) is the sample size. For this problem, \(p' = 315 / 1500 = 0.21\), \(p_0 = 0.18\) and \(n = 1500\). Substituting these values into the formula, we can calculate the z-score.
03

Calculation of P-value

The p-value is the probability of observing a more extreme test statistic in the direction of the alternative hypothesis while the null hypothesis is true. It is calculated using the standard normal (Z) distribution table (or using statistical software). The p-value is found by looking up the computed Z in a Z-distribution table.
04

Compare P-Value and Significance Level

After the p-value is calculated, it is compared with the significance level (\(伪\) value) which is 0.05 in this scenario. If the p-value is less than the significance level, reject the null hypothesis.
05

Conclusion

If the null hypothesis is rejected, then it can be concluded based on the sample data at 5% significance level, that the proportion of shoppers who currently prefer to buy generic instead of name-brand products is higher than 0.18. If not rejected, we cannot conclude that the proportion has increased.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement you aim to test. It assumes no change or no effect, serving as a default or starting assumption. It is denoted by \(H_0\). In our exercise's context, it states that the proportion of shoppers preferring generic over name-brand products is still 18%, represented as \(H_0: p = 0.18\). This means that, according to the null hypothesis, there is no increase from the previously reported 18% preference rate.
It's important because it's what you test against to see if there's sufficient evidence for change. If your test results show significant evidence against it, you may reject it in favor of the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis is what you propose if you find the null hypothesis unlikely based on your data. It represents a specific effect or change you suspect might be true. This hypothesis is represented by \(H_a\) and is often what you're trying to prove.
In our example, the alternative hypothesis is that more than 18% of shoppers prefer generic products now. It鈥檚 written as \(H_a: p > 0.18\). If your results show a greater proportion than 18%, and it鈥檚 statistically significant, you would consider the alternative hypothesis supported.
Significance Level
The significance level, represented by \(\alpha\), is a threshold set before conducting a test. It represents the probability of rejecting the null hypothesis when it is actually true, known as type I error. For most studies, including our exercise, a 5% significance level, or 0.05, is common. This means you are willing to accept a 5% chance of wrongly rejecting the null hypothesis.
A lower significance level means stricter criteria for evidence, reducing the chances of type I error but potentially increasing the risk of failing to detect a true effect (type II error). In our exercise, if the p-value is below 0.05, it suggests strong evidence against the null hypothesis.
P-Value
The p-value is a critical outcome of your hypothesis test. It tells how likely it is to obtain your test results, or more extreme, assuming the null hypothesis is true. A smaller p-value implies stronger evidence against the null hypothesis.
In our scenario, after calculating the test statistic, you use it to find the p-value. If this p-value is lower than the significance level (0.05), you reject the null hypothesis. The p-value helps you decide if the observed data differ enough from what's expected under the null hypothesis, guiding you towards the alternative hypothesis.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. For comparing proportions like in our exercise, the Z-score (standardized score) is commonly used.
The formula for the test statistic in our problem is:
  • \(z = \frac{p' - p_0}{\sqrt{p_0(1 - p_0) / n}}\)
where \(p'\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size.
This statistic allows us to convert our hypothesis into a problem regarding standard deviations from an expected mean, facilitating the calculation of the critical p-value.

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Most popular questions from this chapter

According to a 2014 CIRP Your First College Year Survey, \(88 \%\) of the first- year college students said that their college experience exposed them to diverse opinions, cultures, and values (www. heri.ucla.edu). Suppose in a recent poll of 1800 first-year college students, \(91 \%\) said that their college experience exposed them to diverse opinions, cultures, and values. Perform a hypothesis test to determine if it is reasonable to conclude that the current percentage of all firstyear college students who will say that their college experience exposed them to diverse opinions, cultures, and values is higher than \(88 \% .\) Use a \(2 \%\) significance level, and use both the \(p\) -value and the critical-value approaches.

A study claims that \(65 \%\) of students at all colleges and universities hold off-campus (part-time or full-time) jobs. You want to check if the percentage of students at your school who hold off-campus jobs is different from \(65 \%\). Briefly explain how you would conduct such a test. Collect data from 40 students at your school on whether or not they hold off-campus jobs. Then, calculate the proportion of students in this sample who hold off-campus jobs. Using this information, test the hypothesis. Select your own significance level.

According to a Bureau of Labor Statistics release of February \(20,2015,79 \%\) of American children under age 18 lived with at least one other sibling in 2014 . Suppose that in a recent sample of 2000 American children under age 18,1620 were living with at least one other sibling. a. Using the critical-value approach and \(\alpha=.05\), test if the current percentage of all American children under age 18 who live with at least one other sibling is different from \(79 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.05 ?\)

A random sample of 500 observations produced a sample proportion equal to \(.38\). Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.05\). a. \(H_{0}=p=.30\) versus \(H_{1}: p>.30\) b. \(H_{0^{-}} p=.30\) versus \(\quad H_{1}: p \neq .30\)

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using a \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95\) ? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter 3 . Then make the test of hypothesis about \(\mu .\) )
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