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Chapter 11: Problem 1

Perform the appropriate hypothesis test. A random sample of \(n_{1}=120\) individuals results in \(x_{1}=43\) successes. An independent sample of \(n_{2}=130\) individuals results in \(x_{2}=56\) successes. Does this represent sufficient evidence to conclude that \(p_{1} \neq p_{2}\) at the \(\alpha=0.01\) level of significance?

Short Answer

Expert verified
Fail to reject the null hypothesis. There is insufficient evidence at the \( \alpha=0.01 \) level to conclude that \ p_{1} \ and \ p_{2} \ are different.

Step by step solution

01

Define the hypotheses

The null hypothesis \(\text{H}_{0}\) is that the population proportions are equal: \(\text{H}_{0}: p_{1} = p_{2}\). The alternative hypothesis \(\text{H}_{1}\) is that the population proportions are not equal: \(\text{H}_{1}: p_{1} eq p_{2}\).
02

Calculate the sample proportions

Calculate sample proportions for both groups:\(\hat{p}_1 = \frac{x_1}{n_1} = \frac{43}{120} = 0.358\) \(\hat{p}_2 = \frac{56}{130} = 0.431\)
03

Calculate the pooled sample proportion

The pooled sample proportion \(\hat{p}\) is calculated as \(\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{43 + 56}{120 + 130} = \frac{99}{250} = 0.396\)
04

Calculate the standard error

The standard error (SE) for the difference in sample proportions is: \[SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.396(1 - 0.396)(\frac{1}{120} + \frac{1}{130})} = \sqrt{0.396 \times 0.604 \times (0.00833 + 0.00769)} = \sqrt{0.0047} \approx 0.068\]
05

Calculate the test statistic

The test statistic (z) is calculated by: \[z = \frac{(\hat{p}_1 - \hat{p}_2)}{SE} = \frac{(0.358 - 0.431)}{0.068} \approx -1.07\]
06

Determine the critical value and make a decision

For a two-tailed test with \(\text{H}_{1} : p_{1} eq p_{2}\) at \(\text{\alpha} = 0.01\), the critical z-value is \(\text{Z}_{\alpha/2} \approx \pm 2.576\). Since the calculated z-value \(-1.07\) does not fall outside the critical values of \(-2.576\) and \(+2.576\), we fail to reject the null hypothesis.

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Key Concepts

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

Population Proportions
Population proportions refer to the percentage of individuals in a population that have a particular characteristic. For example, if we are studying the proportion of people who prefer a certain brand, and we have 100 people in our sample, with 40 preferring the brand, the sample proportion is 40%.

In hypothesis testing involving population proportions, you are often given two sample proportions. You compare these to see if there is a significant difference between them.

In the provided exercise, the sample sizes are 120 and 130, with successes of 43 and 56 respectively. By calculating the sample proportions for each group, \(\hat{p}_1 = 0.358\) and \(\hat{p}_2 = 0.431\), we can progress towards understanding if the population proportions \(p_1 eq p_2 \). This sets us up for the pooled sample proportion calculation.
Pooled Sample Proportion
The pooled sample proportion combines successes and sample sizes from different groups to get an overall proportion. This is especially useful in hypothesis testing when comparing two proportions.

In our exercise, we combined the successes from both samples (43 and 56) and the total sample sizes (120 and 130) to figure out the pooled sample proportion \( \hat{p} \).

The formula for the pooled sample proportion is:
\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{99}{250} = 0.396 \]

Why is this necessary? Using a pooled sample proportion allows us to calculate the standard error, an essential step before determining the test statistic in hypothesis testing. The pooled proportion helps assume a common probability of success between the two groups which simplifies the testing process.
Standard Error
Standard Error (SE) measures the variability or spread of the sample proportion estimates. It gives an idea of how much the sample proportion differs from the true population proportion.

In hypothesis testing for comparing two population proportions, the SE helps quantify the expected difference between sample proportions if the null hypothesis is true (i.e., no real difference between populations).

The formula used in our exercise for calculating the standard error is:
\[SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} \]

From our data, this is:
\[SE = \sqrt{0.396(1 - 0.396)(\frac{1}{120} + \frac{1}{130})} = \sqrt{0.396 \times 0.604 \times 0.016} = \sqrt{0.0047} \approx 0.068 \]

This SE value is then used to calculate the z-test statistic, which helps determine whether to reject the null hypothesis. The smaller the standard error, the more confident we can be in our sample statistic.

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Most popular questions from this chapter

A Secchi disk is an 8 -inch-diameter weighted disk that is painted black and white and attached to a rope. The disk is lowered into water and the depth (in inches) at which it is no longer visible is recorded. The measurement is an indication of water clarity. An environmental biologist interested in determining whether the water clarity of the lake at Joliet Junior College is improving takes measurements at the same location on eight dates during the course of a year and repeats the measurements on the same dates five years later. She obtains the following results: $$ \begin{array}{lcccccccc} \text { Observation } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \text { Date } & \mathbf{5 / 1 1} & \mathbf{6 / 7} & \mathbf{6 / 2 4} & \mathbf{7 / 8} & \mathbf{7 / 2 7} & \mathbf{8 / 3 1} & 9 / 30 & \mathbf{1 0 / 1 2} \\ \hline \begin{array}{l} \text { Initial } \\ \text { depth, } X_{i} \end{array} & 38 & 58 & 65 & 74 & 56 & 36 & 56 & 52 \\ \hline \begin{array}{l} \text { Depth five } \\ \text { years later, } Y_{i} \end{array} & 52 & 60 & 72 & 72 & 54 & 48 & 58 & 60 \\ \hline \end{array} $$ (a) Why is it important to take the measurements on the same date? (b) Does the evidence suggest that the clarity of the lake is improving at the \(\alpha=0.05\) level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. (c) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (b)?

Determine whether the sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitative. The Gallup Organization asked 1050 randomly selected adult Americans age 18 or older who consider themselves to be religious, 鈥淒o you believe it is morally acceptable or morally wrong [rotated] to conduct medical research using stem cells obtained from human embryos?鈥� The same question was asked to 1050 randomly selected adult Americans age 18 or older who do not consider themselves to be religious. The goal of the study was to determine whether the proportion of religious adult Americans who believe it is morally wrong to conduct medical research using stem cells obtained from human embryos differed from the proportion of non-religious adult Americans who believe it is morally wrong to conduct medical research using stem cells obtained from human embryos.

In Problems 13-16, construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=368, n_{1}=541, x_{2}=421, n_{2}=593,90 \%\) confidence

Clifford Adelman, a researcher with the Department of Education, followed a cohort of students who graduated from high school in \(1992,\) monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to compare students who first attended a community college to those who immediately attended and remained at a four-year institution. The sample standard deviation time to complete a bachelor's degree of the 268 students who transferred to a four-year school after attending community college was \(1.162 .\) The sample standard deviation time to complete a bachelor's degree of the 1145 students who immediately attended and remained at a four-year institution was \(1.015 .\) Assuming the time to earn a bachelor's degree is normally distributed, does the evidence suggest the standard deviation time to earn a bachelor's degree is different between the two groups? Use the \(\alpha=0.05\) level of significance.

Assume that the populations are normally distributed. (a) Test whether \(\mu_{1} \neq \mu_{2}\) at the \(\alpha=0.05\) level of significance for the given sample data. (b) Construct a \(95 \%\) confidence interval about \(\mu_{1}-\mu_{2}\). $$ \begin{array}{ccc} & \text { Sample } \mathbf{1} & \text { Sample } 2 \\ \hline n & 20 & 20 \\ \hline \bar{x} & 111 & 104 \\ \hline s & 8.6 & 9.2 \\ \hline \end{array} $$
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