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Chapter 16: Problem 611

A sample of Democrats and a sample of Republicans were polled on an issue. Of 200 Republicans, 90 would vote yes on the issue; of 100 Democrats, 58 would vote yes. Can we say that more Democrats than Republicans favor the issue at the \(1 \%\) level of significance?

Short Answer

Expert verified
We used hypothesis testing for comparing two proportions to determine if a higher proportion of Democrats would vote yes compared to Republicans. The calculated p-value (0.0110) is greater than the level of significance (0.01), so we fail to reject the null hypothesis. Thus, we cannot say that more Democrats than Republicans favor the issue at the 1% level of significance.

Step by step solution

01

Set up the null and alternative hypotheses

The null hypothesis (H0) states that there is no difference in the proportion of Democrats and Republicans who would vote yes. The alternative hypothesis (H1) states that a higher proportion of Democrats would vote yes. H0: p_D - p_R = 0 H1: p_D - p_R > 0 where p_D is the proportion of Democrats who would vote yes and p_R is the proportion of Republicans who would vote yes.
02

Calculate sample proportions and their difference

First, let's calculate the sample proportions for each group: Democrats: \(\hat{p}_D = \frac{58}{100} = 0.58\) \\ Republicans: \(\hat{p}_R = \frac{90}{200} = 0.45\) Calculate the difference between the sample proportions: \(\hat{p}_D - \hat{p}_R = 0.58 - 0.45 = 0.13\)
03

Calculate the pooled proportion and standard error

The pooled proportion is calculated as follows: \(p_{pool} = \frac{(58 + 90)}{(100 + 200)} = \frac{148}{300} \approx 0.4933\) The standard error is calculated using the pooled proportion: \(SE = \sqrt{\frac{p_{pool}(1-p_{pool})}{n_D} + \frac{p_{pool}(1-p_{pool})}{n_R}} = \sqrt{\frac{0.4933(1-0.4933)}{100} + \frac{0.4933(1-0.4933)}{200}} \approx 0.0567\)
04

Calculate the z-score and the p-value

The z-score is calculated as follows: \(z = \frac{(\hat{p}_D - \hat{p}_R) - 0}{SE} = \frac{0.13 - 0}{0.0567} \approx 2.29\) Using a standard normal distribution table, find the p-value corresponding to the z-score: p-value = P(Z > 2.29) 鈮� 0.0110
05

Compare the p-value to the level of significance

Now, we compare the calculated p-value (0.0110) to the level of significance (0.01): 0.0110 > 0.01 Since the p-value is greater than the level of significance, we fail to reject the null hypothesis. Therefore, we cannot say that more Democrats than Republicans favor the issue at the 1% level of significance.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding null and alternative hypotheses is crucial when conducting hypothesis testing. The null hypothesis, often denoted as H0, represents a statement of no effect or no difference - in the context of the given exercise, it suggests that the proportion of Democrats (\( p_D \)) and Republicans (\( p_R \)) who would vote 'yes' is the same.

On the other hand, the alternative hypothesis, denoted as H1 or Ha, proposes the opposite - here, it signifies that the proportion of Democrats voting 'yes' is greater than that of Republicans. These starting points set the stage for testing and potential rejection based on the evidence from the sample data.
Pooled Proportion
The pooled proportion is a weighted average that combines the sample proportions when the two samples are assumed under the null hypothesis to have the same proportion. In hypothesis testing regarding two populations, this allows for a more accurate estimate of the shared proportion, particularly when comparing differences.

It is calculated by dividing the total number of successes (for instance, 'yes' votes) by the total number of observations in both samples. The provided problem combines the number of affirmative responses from both Democrats and Republicans, dividing by the total number of individuals surveyed to produce the pooled proportion.
Standard Error
The standard error (SE), a crucial component in hypothesis testing, measures the variability in sample statistics - such as the difference in proportions in our case - from sample to sample. It is different from standard deviation, which measures the variability within a single sample.

In the context of proportional differences between two independent samples, it factors in the pooled proportion and the sample sizes, yielding a gauge of how much the observed differences could vary due to random chance alone. The formula uses the pooled proportion to represent the common proportion from both groups under the null hypothesis.
Z-score
To determine how extreme the observed results are under the null hypothesis, we calculate the z-score. This score tells us how many standard errors our sample's difference in proportion is away from the hypothesized difference (usually zero).

The z-score in the given exercise is a standardized value that translates the observed difference into the language of standard deviation units. A high absolute value of the z-score indicates that the observed difference is not likely the result of random variation alone and hence might be significant.
P-value
One of the most pivotal concepts in hypothesis testing is the p-value. This value helps us decide whether the observed data are extreme enough to reject the null hypothesis.

It is the probability of obtaining a result at least as extreme as the one we've observed, assuming that the null hypothesis is true. If the p-value is low, it suggests that the observed result is quite unlikely under the null hypothesis, casting doubt on its validity and potentially pointing to the alternative being true.
Level of Significance
Lastly, the level of significance serves as the benchmark for determining the strength of evidence against the null hypothesis required to reject it. Typically denoted as alpha (\( \textstyle \(\alpha\) \)), it's pre-set by the researcher, with common levels being 0.05, 0.01, or 0.10.

In the exercise, a 1% level of significance (\( \textstyle \(\alpha = 0.01\) \) means that there's less than a 1% chance the researcher is willing to wrongly reject the null hypothesis. If the p-value is less than the specified alpha level, it indicates there's sufficient evidence to reject the null hypothesis in favor of the alternative.

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Most popular questions from this chapter

You are studying problem solving performance using time as a measure. Your subjects usually either solve the problems quickly or take a long time, but only rarely do they take an intermediate amount of time. You select a sample of 10 subjects and give them training that you believe will reduce their mean time for solving the problem set. You wish to compare their performance with that of another sample of 10 who did not receive the instruction. Outline an appropriate statistical test and suggest changes in the plan of the experiment. Use a level of significance of \(.05\).

Consider the random variable \(\mathrm{X}\) which has a binomial distribution with \(\mathrm{n}=5\) and the probability of success on a single trial, \(\theta\). Let \(\mathrm{f}(\mathrm{x} ; \theta)\) denote the probability distribution function of \(\mathrm{X}\) and let \(\mathrm{H}_{0}: \theta=1 / 2\) and \(\mathrm{H}_{1}: \theta=3 / 4\). Let the level of significance \(\alpha=1 / 32\). Determine the best critical region for the test of the null hypothesis \(\mathrm{H}_{0}\) against the alternate hypothesis \(\mathrm{H}_{1}\). Do the same for \(\alpha=6 / 32\).

Suppose we have a type of battery for which we take a sample of 10 batteries. The mean operating life of these batteries is \(18.0\) hours with a standard deviation of \(3.0\) hours. Suppose also that we have a new type of battery for which we take a sample of 17 batteries. The mean of this sample is \(22.0\) hours with a standard deviation of \(6.0\) hours. Determine for a \(1 \%\) level of significance whether there is a significant difference between the means of the two samples. Also determine at a \(1 \%\) level of significance whether we can conclude that the new batteries are superior to the old ones.

Suppose that you want to decide which of two equally-priced brands of light bulbs lasts longer. You choose a random sample of 100 bulbs of each brand and find that brand \(\mathrm{A}\) has sample mean of 1180 hours and sample standard deviation of 120 hours, and that brand \(\mathrm{B}\) has sample mean of 1160 hours and sample standard deviation of 40 hours. What decision should you make at the \(5 \%\) significance level?

In \(1964,40 \%\) of shipments of U.S. cotton to Germany were sent to arbitration because of complaints about the quality being substandard, according to Time Magazine. Would it signify a real worsening of the situation if 20 out of the first 40 shipments in 1965 were likewise the cause of complaint, or might this difference be attributed to chance?
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