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I want red! A candy maker offers Child and Adult bags of jelly beans with different color mixes. The company claims that the Child mix has 30% red jelly beans while the Adult mix contains 15% red jelly beans. Assume that the candy maker鈥檚 claim is true. Suppose we take a random sample of 50 jelly beans from the Child mix and a separate random sample of 100 jelly beans from the Adult mix. (a) Find the probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample. Show your work. (b) Suppose that the Child and Adult samples contain an equal proportion of red jelly beans. Based on your result in part (a), would this give you reason to doubt the company鈥檚 claim? Explain.

Short Answer

Expert verified
(a) P(\(\hat{p}_1 \leq \hat{p}_2\)) = 0.0207. (b) Yes, equal proportions would suggest the company's claim might be inaccurate.

Step by step solution

01

Identify Given Information

The problem states that in the Child mix, 30% of the jelly beans are red, and in the Adult mix, 15% are red. We have a sample size of 50 for the Child mix and 100 for the Adult mix. Let \( p_1 = 0.30 \) and \( n_1 = 50 \) for Child, and \( p_2 = 0.15 \) and \( n_2 = 100 \) for Adult.
02

Determine Sample Statistics

Calculate the expected number of red jelly beans and standard deviations for both samples: - Child mix: Expected proportion \( \hat{p}_1 = p_1 = 0.30\). Standard deviation \( \sigma_1 = \sqrt{\frac{p_1(1-p_1)}{n_1}} = \sqrt{\frac{0.30 \times 0.70}{50}} \approx 0.0648 \).- Adult mix: Expected proportion \( \hat{p}_2 = p_2 = 0.15 \). Standard deviation \( \sigma_2 = \sqrt{\frac{p_2(1-p_2)}{n_2}} = \sqrt{\frac{0.15 \times 0.85}{100}} \approx 0.0357 \).
03

Set Up the Hypotheses

We are asked to find \( P(\hat{p}_1 \leq \hat{p}_2) \). This compares the sampled proportions to ensure the Child proportion is less than or equal to the Adult proportion.
04

Calculate Z-score for Differences in Proportions

Find the standard deviation of the difference between the proportions: \( \sigma_\text{difference} = \sqrt{\sigma_1^2 + \sigma_2^2} = \sqrt{0.0648^2 + 0.0357^2} \approx 0.0736 \).Calculate the difference in sample proportions: \( \hat{p}_1 - \hat{p}_2 = 0.30 - 0.15 = 0.15 \). Now, calculate the Z-score: \( Z = \frac{0.15 - 0}{0.0736} \approx 2.037 \).
05

Find the Probability

Using the Z-score table, find the probability that corresponds to \( Z = 2.037 \). This value is approximately 0.9793. Therefore, \( P(\hat{p}_1 \leq \hat{p}_2) = 1 - 0.9793 = 0.0207 \).
06

Interpret the Result

With \( P(\hat{p}_1 \leq \hat{p}_2) = 0.0207 \), there's only a 2.07% chance that the Child mix sample will have a smaller or equal proportion of red jelly beans than the Adult mix.
07

Answer Part (b): Evaluate the Claim

Since the probability is very low (2.07%), observing equal proportions or more red jelly beans in the Adult mix sample than in the Child mix sample provides strong evidence against the company's claim. Hence, if the proportions were equal, it would be reasonable to doubt the claim.

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Key Concepts

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

Understanding Sampling Distributions
Sampling distributions allow us to understand how sample statistics behave when drawn from a population. In our exercise, we were dealing with the proportions of red jelly beans in two different samples鈥攖he Child and Adult mixes. When we take multiple samples, each one will have its own sample proportion of red jelly beans, like 30% for the Child mix and 15% for the Adult mix. This distribution of sample proportions will center around the true population proportion (the company's claims), but there will always be some variability.

We use the standard deviation of these sample proportions (also known as the standard error) to measure this variability. It helps predict how much the sample statistic (like the proportion of red beans) will deviate from the true population proportion. Calculating standard deviations for both Child and Adult mixes, using formulas like \[ \sigma = \sqrt{\frac{p(1-p)}{n}} \] is key in making any statistical inferences.
Basics of Hypothesis Testing
Hypothesis testing is a fundamental part of statistics used to make inferences about population parameters. In the given exercise, we set up a hypothesis to determine if there's substantial evidence against the company's claim about the proportion of red jelly beans. The null hypothesis usually states no effect or no difference鈥攊n this case, that the Child have equal or lesser proportion as the Adult mix.

By calculating our test statistic, which follows a normal distribution in this scenario, we could compare against critical values. This type of statistical testing helps us understand if our observed results (i.e., the sample proportions) can be attributed to chance, or if it suggests a systemic difference. If our observed samples have a very low probability under the null hypothesis, it implies a potential discrepancy from the claim, leading us to question its validity.
Understanding Proportions in Statistics
Proportions are a way of expressing part of a whole, and in this case, they relate to the percentage of red jelly beans within each sample. Specifically, we have proportions of 0.30 and 0.15 for the Child and Adult mixes, respectively.

Understanding how these proportions are used in the context of sampling helps us assess probabilities across different groups. Calculating these involves dividing the number of red jelly beans observed in a sample by the total number of jelly beans in that sample.

By comparing the proportions from our samples, we were able to compute the likelihood (using a Z-score) that the Child mix proportion would be less than or equal to the Adult mix one. If the probability of the observed proportions occurring under the claimed condition is extremely low, it suggests that the initial claim about proportions may not be accurate.
Role of Standard Deviation in Inference
Standard deviation plays a pivotal role in statistical inference as it offers a numerical measure of spread or variability within a dataset. In our candy jelly bean example, it helped us understand how much the sample proportions could vary from the true population proportion鈥攅ssentially quantifying uncertainty.

Calculating the standard deviation involved using the binomial standard deviation formula, which translates into:
\[ \sigma = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the sample proportion and \( n \) is the sample size. The combined standard deviation of the difference in our jelly bean proportions for two samples was determined to bring out a clearer picture when measuring a change or verifying a claim.

This understanding directly influences decisions in hypothesis testing, where standard deviation is a crucial part of calculating Z-scores and other test statistics used to evaluate the differences observed between groups.

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Most popular questions from this chapter

Paying for college College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender. Here are the data in summary form:\(^{33}\) $$\begin{array}{llll}{\text { Group }} & {n} & {\overline{x}} & {s_{x}} \\\ \hline \text { Males } & {675} & {\$ 1884.52} & {\$ 13688.37} \\ {\text { Females }} & {621} & {\$ 1360.39} & {\$ 1037.46}\end{array}$$ (a) How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? A graph of the data reveals no outliers. The use of two-sample t procedures is still justified. Why? (b) Construct and interpret a 90% confidence interval for the difference between the mean summer earnings of male and female students at this university. (c) Interpret the 90% confidence level in the context of this study.

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What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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