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Chapter 10: Problem 19

Children make choices Many new products introduced into the market are targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different age groups were compared on their ability to sort new products into the correct product category (milk or juice). \({ }^{14}\) Here are some of the data: $$ \begin{array}{lll} \hline \text { Age group } & N & \text { Number who sorted correctly } \\ \text { 4- to 5-year-olds } & 50 & 10 \\ \text { 6- to 7-year-olds } & 53 & 28 \\ \hline \end{array} $$ Did a significantly higher proportion of the 6 - to 7-year-olds than the 4 - to 5 -year-olds sort correctly? Give appropriate evidence to justify your answer.

Short Answer

Expert verified
Yes, 6- to 7-year-olds sorted more correctly, supported by a z-score of 3.447, exceeding the critical value 1.645.

Step by step solution

01

Define the Hypotheses

To determine if a significantly higher proportion of 6- to 7-year-olds sorted correctly, we set up our null and alternative hypotheses.- Null Hypothesis \((H_0)\): The proportions of 6- to 7-year-olds and 4- to 5-year-olds who sorted correctly are the same. - Alternative Hypothesis \((H_a)\): The proportion of 6- to 7-year-olds who sorted correctly is higher than that of 4- to 5-year-olds.
02

Calculate Sample Proportions

Compute the sample proportions for each age group.- For 4- to 5-year-olds: \( \hat{p}_1 = \frac{10}{50} = 0.2 \)- For 6- to 7-year-olds: \( \hat{p}_2 = \frac{28}{53} \approx 0.528 \)
03

Calculate the Standard Error

Find the pooled proportion and the standard error.- Pooled proportion \( \hat{p} = \frac{10+28}{50+53} = \frac{38}{103} \approx 0.369 \)- Calculate the standard error (SE): \[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.369(0.631)\left(\frac{1}{50} + \frac{1}{53}\right)} \approx 0.095 \]
04

Calculate the Test Statistic

Use the standard error to compute the test statistic (z-score):\[ z = \frac{\hat{p}_2 - \hat{p}_1}{SE} = \frac{0.528 - 0.2}{0.095} \approx 3.447 \]
05

Determine Significance

Compare the calculated z-score to a critical value or use a p-value approach. - Using a significance level of 0.05 for a one-tailed test, the critical z-value is approximately 1.645. - Since the calculated z (3.447) is greater than 1.645, we reject the null hypothesis.

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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 (denoted as \(H_0\)) is a starting point for your study. It is a statement that indicates no effect or no difference in the situation being analyzed. In our context, the null hypothesis asserts that the proportions of children, aged 4-5 and 6-7, who can correctly sort products into categories, are equal. This suggests that age does not influence their ability to sort.

The Null Hypothesis is essential because it sets a baseline for comparison. Without this, we wouldn't know whether any observed differences between groups are genuinely significant or simply due to random chance.

Key things to remember about the Null Hypothesis:
  • It is a statement of no effect or no difference.
  • It is typically accompanied by an "equals" sign (for instance, \(p_1 = p_2\)).
  • The aim of many tests is to determine if we can reject the null hypothesis with confidence.
Alternative Hypothesis
The Alternative Hypothesis (denoted as \(H_a\)) is the statement we are testing against the null hypothesis. It represents what researchers aim to prove or expect to find based on their investigation. In this exercise, the alternative hypothesis states that the proportion of 6- to 7-year-olds who sorted correctly is higher than for the 4- to 5-year-olds.

This hypothesis implies that age might play a role in children's product sorting ability. When we observe data, the goal is to find enough evidence to support this hypothesis, consequently rejecting the null hypothesis.

Important notes on the Alternative Hypothesis include:
  • It suggests there is an effect or a difference, contrary to \(H_0\).
  • In our exercise, it is written as a one-sided inequality: \(p_2 > p_1\).
  • The strength of evidence needed to support \(H_a\) depends on the pre-set significance level of the test (commonly 0.05).
Standard Error
Standard Error (SE) is a measure that tells us how much sample proportions are expected to vary from the true population proportion due to random sampling. It helps us understand the variability in our sample data.

In the context of comparing proportions, such as in our exercise, SE gives us an idea of how reliable our observed difference between the two sample proportions is. The formula for standard error when comparing two proportions is:

\[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]

Where \( \hat{p} \) is the pooled sample proportion, combining both age groups.

Some critical points about Standard Error include:
  • It helps assess the precision of a sample estimate.
  • A smaller SE indicates a more precise estimate of the population parameter.
  • SE is foundational in estimating confidence intervals and calculating test statistics like the z-score.
Z-Score
The Z-Score is a statistic that tells us how many standard errors our observed sample proportion difference is from the null hypothesis value, usually zero in tests comparing two proportions. It serves as the test statistic in hypothesis testing.

In our exercise, the z-score is calculated to understand the significance of the difference between the sorting proficiency of the two age groups. The formula for the z-score here is:

\[ z = \frac{\hat{p}_2 - \hat{p}_1}{SE} \]

Where \( \hat{p}_2 \) and \( \hat{p}_1 \) are the sample proportions for the two groups, and SE is the standard error.

Key points to remember about the Z-Score include:
  • A higher absolute z-score indicates a more significant deviation from the null hypothesis.
  • If the calculated z-score exceeds the critical z-value at a given significance level, the null hypothesis is rejected.
  • The z-score helps standardize differences to see if they are unlikely under the null hypothesis鈥檚 assumption.

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Most popular questions from this chapter

Steroids in high school \(A\) study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance. \({ }^{13}\) Do the data give convincing evidence of a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids? State appropriate hypotheses for a test to answer this question. Define any parameters you use..

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