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91Ó°ÊÓ

Consider the following information obtained from two independent samples: $$ n_{1}=300, \quad \hat{p}_{1}=.55, \quad n_{2}=200, \quad \hat{p}_{2}=.62 $$ Test at a \(1 \%\) significance level if \(p_{1}\) is less than \(p_{2}\).

Short Answer

Expert verified
Based on the test statistic and critical value, we cannot conclude that \(p_{1}\) is less than \(p_{2}\) at a 1% significance level. Therefore, we fail to reject the null hypothesis.

Step by step solution

01

Set up the hypotheses

The null hypothesis is \(H_0: p_{1} - p_{2} = 0\), and the alternative hypothesis is \(H_A: p_{1} - p_{2} < 0\), meaning we are testing if proportion \(p_{1}\) is less than proportion \(p_{2}\).
02

Compute the pooled sample proportion

The pooled sample proportion \(p\) is calculated as \((x_{1} + x_{2}) / (n_{1} + n_{2})\), where \(x_{1} = n_{1}*p_{1}\) and \(x_{2} = n_{2}*p_{2}\). Substituting the given values, \(p = (300*0.55 + 200*0.62) / (300 + 200) = 0.575.\)
03

Compute the standard error

The standard error (SE) is \(\sqrt{[p*(1-p )] * [(1/n_{1}) + (1/n_{2})]}\). Plugging in the calculated value of \(p\) and the given values for \(n_{1}\) and \(n_{2}\), we get \(SE = \sqrt{(0.575*0.425) * [(1/300) + (1/200)]} = 0.033.\)
04

Compute the test statistic (Z score)

The test statistic (Z) is \((p_{1} - p_{2} - 0) / SE\). Substituting the standard error we computed previously, and the given values for \(p_{1}\) and \(p_{2}\), we obtain \(Z = (0.55 - 0.62) / 0.033 \approx -2.121.\)
05

Find the critical value and perform the test

A 1% significance level means a confidence level of 99%. Looking up Z tables for a one-sided test at 99% confidence, we find a critical value of approximately -2.33. Because our test statistic (-2.121) is greater than the critical value (-2.33), 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.

Proportions
In hypothesis testing, proportions are an essential metric often needed to understand how two datasets compare. A proportion in statistics represents a part of the whole. It's essentially a percentage that shows how many parts of the data set represent some characteristic. For example, if we consider a sample of voters and find that 55% support a candidate, then this proportion, denoted as \( \hat{p} \), is 0.55.

In our problem, we have two proportions: \( \hat{p}_1 = 0.55 \) and \( \hat{p}_2 = 0.62 \). These represent the proportions of the characteristic of interest in samples 1 and 2. This is crucial for determining if these samples come from populations with the same or different proportions.
  • Proportions are calculated by dividing the number of favorable outcomes by the total number of outcomes.
  • The differences in proportions are then analyzed to draw conclusions about the populations.
Sample Size
Sample size, denoted as \( n \), is the number of observations in a sample. It is a critical factor in hypothesis testing as it can influence the reliability and validity of the results. Larger sample sizes tend to provide more reliable estimates of population parameters.

In our example, we have two sample sizes: \( n_1 = 300 \) and \( n_2 = 200 \). Each comes from a different population. The larger the sample size, the more accurate the estimate of the population parameter, assuming random sampling.
  • A larger sample size reduces the margin of error and increases the confidence in the resulting statistics.
  • The sample size affects the calculation of the standard error, which in turn affects the test statistic.
Significance Level
The significance level, often denoted as \( \alpha \), is the threshold at which we decide whether or not to reject the null hypothesis. It represents the probability of committing a Type I error, which is rejecting a true null hypothesis. Common significance levels are 0.05, 0.01, and 0.10.

In this problem, a 1% level of significance is used (\( \alpha = 0.01 \)). This is a strict threshold, indicating that we need strong evidence against the null hypothesis to reject it. The significance level determines the critical value from the Z distribution:
  • A smaller \( \alpha \) (like 0.01) means we require stronger evidence against the null hypothesis.
  • It corresponds to a confidence level of 99% in this scenario.
Null Hypothesis
The null hypothesis, denoted \( H_0 \), is a statement used in hypothesis testing that assumes no effect or no difference exists. It serves as the default or baseline assumption. In hypothesis tests involving proportions, the null hypothesis usually posits that the two proportions are equal.

In our case, the null hypothesis is \( H_0: p_1 - p_2 = 0 \). This means we assume there is no difference between the two population proportions. The goal of the test is to see if there is enough evidence to reject this hypothesis.
  • Starting with the null hypothesis simplifies the testing process by providing a standard to test against.
  • The alternative hypothesis (\( H_A \)), in contrast, suggests what we suspect might be true.
Test Statistic
The test statistic is a standardized value calculated from sample data during a hypothesis test. It indicates how far the sample statistic is from the null hypothesis if the null is true. In tests of proportions, the test statistic under null hypothesis is often in the form of a Z-score, especially for large samples.

For this exercise, we have calculated the test statistic as \( Z = -2.121 \). This was obtained by taking the difference of proportions and dividing by the standard error:
  • The test statistic helps us determine how extreme the sample results are under the assumption that the null hypothesis is true.
  • It is then compared to a critical value to decide whether we have enough evidence to reject the null hypothesis.

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Most popular questions from this chapter

We wish to estimate the difference between the mean scores on a standardized test of students taught by Instructors \(\mathrm{A}\) and \(\mathrm{B}\). The scores of all students taught by Instructor A have a normal distribution with a standard deviation of 15, and the scores of all students taught by Instructor \(\mathrm{B}\) have a normal distribution with a standard deviation of \(10 .\) To estimate the difference between the two means, you decide that the same number of students from each instructor's class should be observed. a. Assuming that the sample size is the same for each instructor's class, how large a sample should be taken from each class to estimate the difference between the mean scores of the two populations to within 5 points with \(90 \%\) confidence? b. Suppose that samples of the size computed in part a will be selected in order to test for the difference between the two population mean scores using a .05 level of significance. How large does the difference between the two sample means have to be for you to conclude that the two population means are different? c. Explain why a paired-samples design would be inappropriate for comparing the scores of Instructor A versus Instructor \(\mathrm{B}\).

In parts of the eastern United States, whitetail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent A, and the other 27 receive repellent \(\mathrm{B}\). For each of the 56 gardens, the time elapsed between application of the repellent and the appearance of the first deer in the garden is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(\mathrm{B}\), the mean time is 92 hours. Assume that the two populations of elapsed times have approximately normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at a \(2 \%\) significance level whether the mean elapsed times for repellents \(\mathrm{A}\) and \(\mathrm{B}\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.

When are the samples considered large enough for the sampling distribution of the difference between two sample proportions to be (approximately) normal?

The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ Test at a \(5 \%\) significance level if \(\mu_{1}\) is less than \(\mu_{2}\).

An economist was interested in studying the impact of the recession of a few years ago on dining out, including drive-through meals at fast-food restaurants. A random sample of 48 families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95 .\) Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average of \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have unknown and unequal population standard deviations. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in dining out spending levels for the two populations. b. Using a \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

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