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Would you favor spending more federal tax money on the arts? This question was asked by a research group on behalf of The National Institute (Reference: Painting by Numbers, J. Wypijewski, University of California Press). Of a random sample of \(n_{1}=93\) politically conservative voters, \(r_{1}=21\) responded yes. Another random sample of \(n_{2}=83\) politically moderate voters showed that \(r_{2}=22\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha=0.05\).

Short Answer

Expert verified
There is not enough evidence to conclude that the proportion of conservatives in favor is less than that of moderates.

Step by step solution

01

Define Hypotheses

To determine whether there is a significant difference between the two population proportions, setup the null and alternative hypotheses. The null hypothesis is that there is no difference between the population proportions of conservatives and moderates, symbolically expressed as \( H_0: p_1 \geq p_2 \). The alternative hypothesis claims that the proportion of conservatives who favor spending more tax money on arts is less than that of moderates, expressed as \( H_a: p_1 < p_2 \).
02

Calculate Sample Proportions

Calculate the sample proportion of each group. For conservatives, \( \hat{p_1} = \frac{r_1}{n_1} = \frac{21}{93} \approx 0.226 \). For moderates, \( \hat{p_2} = \frac{r_2}{n_2} = \frac{22}{83} \approx 0.265 \).
03

Calculate the Pooled Proportion

The pooled sample proportion \( \hat{p} \) is computed by \( \hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{21 + 22}{93 + 83} = \frac{43}{176} \approx 0.244 \).
04

Calculate the Standard Error

Use the pooled sample proportion to calculate the standard error: \( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.244(1-0.244)\left(\frac{1}{93} + \frac{1}{83}\right)} \approx 0.065 \).
05

Compute the Test Statistic

Use the sample proportions and the standard error to compute the test statistic: \( z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.226 - 0.265}{0.065} \approx -0.600 \).
06

Determine the Critical Value and Compare

For a one-tailed test with \( \alpha = 0.05 \), the critical value for \( z \) is approximately \( -1.645 \). Since the calculated \( z \approx -0.600 \) is greater than \( -1.645 \), 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
Understanding population proportions is crucial in hypothesis testing. Proportions represent the ratio or percentage of the population that exhibits a certain characteristic or behavior. In the context of our exercise, the population proportion is the part of voters in each political category who favor spending more tax money on the arts.

For conservatives and moderates, population proportions were estimated using samples. Even though we don't know the whole population's exact figures, we can make educated guesses based on samples we've collected. In hypothesis testing, we're interested in comparing population proportions. We estimate them using sample proportions, computed as the number of specific responses divided by the total number of survey participants in each group.

Sample proportions offer a snapshot of what we suspect the population proportions might look like, enabling us to form null and alternative hypotheses and eventually run meaningful statistical tests.
Sampling Methods
Successful hypothesis testing often begins with effective sampling methods. Sampling is selecting a subset of individuals from a population to estimate characteristics of the whole group. Here, random samples were used from two voter categories: conservatives and moderates.

Random sampling is fundamental in statistics because it helps eliminate bias, making the sample more likely to reflect the population accurately. Each member of the population has an equal chance of being included in the sample, which provides a wider spectrum of opinions and characteristics.

For our exercise:
  • The conservative sample comprised 93 voters.
  • The moderate sample included 83 voters.
These samples serve as a groundwork for accurately computing sample proportions and performing further statistical analyses, such as hypothesis testing.
Statistical Significance
Statistical significance plays a pivotal role in hypothesis testing, as it helps us interpret the results with regard to our hypotheses. When conducting these tests, we want to know whether the observed differences (or similarities) between groups are due to chance or indicate a genuine effect. In the exercise, statistical significance is evaluated using a significance level, \( \alpha = 0.05 \).

The choice of \( \alpha = 0.05 \) signifies a balanced level of stringency, allowing us to reject the null hypothesis if there's less than a 5% probability that the observed difference happened by chance alone.

Through calculation steps, we find the test statistic, which is then compared to a critical value derived from statistical tables. When the test statistic falls within the critical region, it suggests statistical significance; otherwise, it means the observed results do not provide enough evidence to reject the null hypothesis.

In our example, the computed test statistic \(-0.600\) did not fall within the critical region, suggesting the sample evidence wasn't convincing enough to assert that conservatives favor spending less on the arts compared to moderates. The process underscores the importance of using statistical significance to draw data-driven conclusions.

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Most popular questions from this chapter

Unfortunately, arsenic occurs naturally in some ground water (Reference: Union Carbide Technical Report K/UR-1). A mean arsenic level of \(\mu=8.0\) parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of \(\bar{x}=7.2\) ppb arsenic, with \(s=1.9\) ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use \(\alpha=0.01\).

The body weight of a healthy 3 -month-old colt should be about \(\mu=60 \mathrm{~kg}\) (Source: The Merck Veterinary Mamual, a standard reference manual used in most veterinary colleges). (a) If you want to set up a statistical test to challenge the claim that \(\mu=60 \mathrm{~kg}\), what would you use for the null hypothesis \(H_{0} ?\) (b) In Nevada, there are many herds of wild horses. Suppose you want to test the claim that the average weight of a wild Nevada colt ( 3 months old) is less than \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis \(H_{1} ?\) (c) Suppose you want to test the claim that the average weight of such a wild colt is greater than \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis? (d) Suppose you want to test the claim that the average weight of such a wild colt is different from \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the \(P\) -value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.

When conducting a test for the difference of means for two independent populations \(x_{1}\) and \(x_{2}\), what alternate hypothesis would indicate that the mean of the \(x_{2}\) population is larger than that of the \(x_{1}\) population? Express the alternate hypothesis in two ways.

In general, if sample data are such that the null hypothesis is rejected at the \(\alpha=1 \%\) level of significance based on a two-tailed test, is \(H_{0}\) also rejected at the \(\alpha=1 \%\) level of significance for a corresponding one-tailed test? Explain.

Let \(x\) be a random variable that represents the \(\mathrm{pH}\) of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \(x\) distribution is \(\mu=7.4\) (Reference: Merck Manual, a commonly used reference in medical schools and nursing programs). A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that \(\bar{x}=8.1\) with sample standard deviation \(s=1.9 .\) Use a \(5 \%\) level of significance to test the claim that the drug has changed (either way) the mean \(\mathrm{pH}\) level of the blood.

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