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Chapter 11: Problem 4

Perform the appropriate hypothesis test. A random sample of \(n_{1}=135\) individuals results in \(x_{1}=40\) successes. An independent sample of \(n_{2}=150\) individuals results in \(x_{2}=60\) successes. Does this represent sufficient evidence to conclude that \(p_{1}

Short Answer

Expert verified
There is sufficient evidence to conclude that \(p_{1} < p_{2}\) at the 0.05 level of significance.

Step by step solution

01

Define null and alternative hypotheses

Let \(p_{1}\) and \(p_{2}\) be the proportions of successes in populations 1 and 2, respectively. The null hypothesis \(H_0\) states that there is no difference in the proportions, \(H_0: p_{1} = p_{2}\). The alternative hypothesis \(H_a\) is that the proportion in sample 1 is less than that in sample 2, \(H_a: p_{1} < p_{2}\).
02

Calculate sample proportions

Compute the sample proportions for each sample. For sample 1, \(\hat{p}_{1} = \frac{x_{1}}{n_{1}} = \frac{40}{135} \approx 0.296\). For sample 2, \(\hat{p}_{2} = \frac{x_{2}}{n_{2}} = \frac{60}{150} = 0.4\).
03

Pooled proportion

Calculate the pooled proportion \(\hat{p}\) given by \(\hat{p} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}} = \frac{40 + 60}{135 + 150} = \frac{100}{285} \approx 0.3518\).
04

Test statistic calculation

The test statistic for comparing two proportions can be calculated using: \[ z = \frac{\hat{p}_{1} - \hat{p}_{2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_{1}} + \frac{1}{n_{2}})}} \] Substitute the known values: \[ z = \frac{0.296 - 0.4}{\sqrt{0.3518 (1 - 0.3518) (\frac{1}{135} + \frac{1}{150})}} \approx \frac{-0.104}{0.059} \approx -1.76 \]
05

P-value and decision

Since it is a left-tailed test, find the P-value for \(z = -1.76\). Using standard normal distribution tables or a calculator, the P-value is approximately 0.0392. Compare this with the significance level \(\alpha=0.05\). Since 0.0392 < 0.05, 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, we start with a statement called the null hypothesis, denoted as \( H_0 \). It represents the idea that there is no significant difference between the groups you're testing. In our case, the null hypothesis is expressing that the proportion of successes in population 1 (\( p_1 \)) is equal to the proportion of successes in population 2 (\( p_2 \)).

Mathematically, the null hypothesis is formulated as: \[ H_0: p_1 = p_2 \] By assuming this, we can gather evidence to test whether this assumption holds true or not. We'll compare our findings to this baseline assumption.
alternative hypothesis
The alternative hypothesis is what you would believe if the null hypothesis is shown to be unlikely. It is represented by \( H_a \). For our problem, we consider the scenario where the proportion of successes in sample 1 is less than in sample 2.

The mathematical representation is: \[ H_a: p_1 < p_2 \] Here, we are specifically interested in checking whether there is sufficient evidence to support this claim at a significance level of \( \alpha=0.05 \).
pooled proportion
The pooled proportion comes into play when you want to combine results from two samples. It estimates the overall proportion of successes in both samples combined. This is crucial because it provides a single estimate of the proportion under the null hypothesis assumption.

The formula for the pooled proportion (\( \hat{p} \)) is: \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \] Plugging in the numbers from our samples: \[ \hat{p} = \frac{40 + 60}{135 + 150} = \frac{100}{285} \approx 0.3518 \] This estimate combines both samples to give us a single proportion to base our comparisons on.
test statistic calculation
The test statistic helps us quantify the difference between the sample proportions compared to what we would expect under the null hypothesis. It follows a standardized distribution, often the z-distribution for proportions.

The formula for our z-test statistic is: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p} \left(1 - \hat{p}\right) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \] Substituting our values: \[ z = \frac{0.296 - 0.4}{\sqrt{0.3518 \left(1 - 0.3518\right) \left(\frac{1}{135} + \frac{1}{150}\right)}} \approx \frac{-0.104}{0.059} \approx -1.76 \] This test statistic will be used to find the probability (P-value) of observing such a result under the null hypothesis.
P-value
The P-value lets us determine the significance of our test results. It represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true.

For our left-tailed test, we calculate the P-value for z = -1.76. From standard normal distribution tables or using a calculator, we find that the P-value is approximately 0.0392.

Comparing this P-value to our significance level \( \alpha \ = 0.05\), we see that 0.0392 < 0.05, which means we have enough evidence to reject the null hypothesis. Thus, there is sufficient evidence to conclude that the proportion of successes in sample 1 is less than that in sample 2.

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Most popular questions from this chapter

In October 1947, the Gallup organization surveyed 1100 adult Americans and asked, 鈥淎re you a total abstainer from, or do you on occasion consume, alcoholic beverages?鈥� Of the 1100 adults surveyed, 407 indicated that they were total abstainers. In a recent survey, the same question was asked of 1100 adult Americans and 333 indicated that they were total abstainers. Has the proportion of adult Americans who totally abstain from alcohol changed? Use the a = 0.05 level of significance.

Sugary Beverages It has been reported that consumption of sodas and other sugar-sweetened beverages cause excessive weight gain. Researchers conducted a randomized study in which 224 overweight and obese adolescents who regularly consumed sugar-sweetened beverages were randomly assigned to experimental and control groups. The experimental groups received a one-year intervention designed to decrease consumption of sugar-sweetened beverages, with follow-up for an additional year without intervention. The response variable in the study was body mass index (BMI-the weight in kilograms divided by the square of the height in meters). Results of the study appear in the following table. Source: Cara B. Ebbeling, \(\mathrm{PhD}\) and associates, "A Randomized Trial of Sugar-Sweetened Beverages and Adolescent Body Weight" N Engl J Med 2012;367:1407-16. DOI: 10.1056/NEJMoal2Q3388 $$ \begin{array}{lll} & \text { Experimental } & \text { Control } \\ & \text { Group } & \text { Group } \\ & (n=110) & (n=114) \\ \hline \text { Start of Study } & \text { Mean BMI }=30.4 & \text { Mean BMI }=30.1 \\ & \text { Standard Deviation } & \text { Standard Deviation } \\ & \mathrm{BMI}=5.2 & \mathrm{BMI}=4.7 \\ \hline \text { After One Year } & \text { Mean Change in } & \text { Mean Change in } \\ & \mathrm{BMI}=0.06 & \mathrm{BMI}=0.63 \\ & \text { Standard Deviation } & \text { Standard Deviation } \\ & \text { Change in } & \text { Change in } \\ & \mathrm{BMI}=0.20 & \mathrm{BMI}=0.20 \\ \hline \text { After Two Years } & \text { Mean Change in } & \text { Mean Change in } \\ & \mathrm{BMI}=0.71 & \mathrm{BMI}=1.00 \\ & \text { Standard Deviation } & \text { Standard Deviation } \\ & \text { Change in } & \text { Change in } \\ & \mathrm{BMI}=0.28 & \mathrm{BMI}=0.28 \end{array} $$ (a) What type of experimental design is this? (b) What is the response variable? What is the explanatory variable? (c) One aspect of statistical studies is to verify that the subjects in the various treatment groups are similar. Does the sample evidence support the belief that the BMIs of the subjects in the experimental group is not different from the BMIs in the control group at the start of the study? Use an \(\alpha=0.05\) level of significance. (d) One goal of the research was to determine if the change in BMI for the experimental group was less than that for the control group after one year. Conduct the appropriate test to see if the evidence suggests this goal was met. Use an \(\alpha=0.05\) level of significance. What does this result suggest? (e) Does the sample evidence suggest the change in BMI is less for the experimental group than the control group after two years? Use an \(\alpha=0.05\) level of significance. What does this result suggest? (f) To what population do the results of this study apply?

For each study, explain which statistical procedure (estimating a single proportion; estimating a single mean; hypothesis test for a single proportion; hypothesis test for a single mean; hypothesis test or estimation of two proportions, hypothesis test or estimation of two means, dependent or independent) would most likely be used for the research objective given. Assume all model requirements for conducting the appropriate procedure have been satisfied. While exercising by climbing stairs, is it better to take one stair, or two stairs, at a time? Researchers identified 30 volunteers who were asked to climb stairs for two different 15-minute intervals taking both one stair and two stairs at a time. Whether the volunteer did one stair or two stairs first was determined randomly. The goal of the research was to determine if energy expenditure for each exercise routine was different.

On April \(12,1955,\) Dr. Jonas Salk released the results of clinical trials for his vaccine to prevent polio. In these clinical trials, 400,000 children were randomly divided in two groups. The subjects in group 1 (the experimental group) were given the vaccine, while the subjects in group 2 (the control group) were given a placebo. Of the 200,000 children in the experimental group, 33 developed polio. Of the 200,000 children in the control group, 115 developed polio. (a) What type of experimental design is this? (b) What is the response variable? (c) What are the treatments? (d) What is a placebo? (e) Why is such a large number of subjects needed for this study? (f) Does it appear to be the case that the vaccine was effective?

Determine whether the sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitative. A researcher wishes to determine the effects of alcohol on people鈥檚 reaction time to a stimulus. She randomly divides 100 people aged 21 or older into two groups. Group 1 is asked to drink 3 ounces of alcohol, while group 2 drinks a placebo. Both drinks taste the same, so the individuals in the study do not know which group they belong to. Thirty minutes after consuming the drink, the subjects in each group perform a series of tests meant to measure reaction time.
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