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Chapter 8: Problem 17

According to the Federal Poverty Measure \(12 \%\) of the U.S. population lives in poverty. The governor of a certain state believes that the proportion there is lower. In a sample of size 1,550,163 were impoverished according to the federal measure. a. Test whether the true proportion of the state's population that is impoverished is less than \(12 \%,\) at the \(5 \%\) level of significance. 1\. Compute the observed significance of the test.

Short Answer

Expert verified
The state's impoverished proportion is significantly less than 12% at 5% level; p-value is 0.0257.

Step by step solution

01

Identify the Hypotheses

First, we need to identify the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the true proportion of the state's population that is impoverished is equal to \(12\%\), or \( p = 0.12 \). The alternative hypothesis \( H_a \) is that the true proportion is less than \(12\%\), or \( p < 0.12 \).
02

Calculate the Test Statistic

We use the formula for the test statistic for a proportion: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion under \( H_0 \), and \( n \) is the sample size. Here, \( \hat{p} = \frac{163}{1550} \approx 0.105 \), \( p_0 = 0.12 \), and \( n = 1550 \).
03

Calculate the Standard Error

The standard error for the proportion is calculated as: \( \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.12 \times (1-0.12)}{1550}} \approx 0.0077 \).
04

Compute the Test Statistic

Substituting the values into the formula for the z-test statistic, we have: \( z = \frac{0.105 - 0.12}{0.0077} \approx -1.948 \).
05

Determine the Critical Value and Decision

At a \(5\%\) level of significance for a one-tailed test, the critical z-value is approximately \(-1.645\). Since our computed z-value of \(-1.948\) is less than \(-1.645\), we reject the null hypothesis.
06

Calculate the Observed Significance (p-value)

To find the observed significance or p-value, we look up the z-value in standard normal distribution tables, or use a calculator. A z-value of \(-1.948\) corresponds to a p-value of approximately 0.0257.
07

Conclusion

Since the p-value (0.0257) is less than the significance level (0.05), there is sufficient evidence 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.

Proportion Test
The proportion test is a way to determine if there is a significant difference between a sample proportion and a known population proportion. Suppose you want to check if a certain characteristic is more common in one group compared to a known standard. In the example provided, the aim is to test whether less than 12% of the state's population lives in poverty compared to the national rate. Here's how it works:
  • Null Hypothesis ( \( H_0 \) ): Assumes no change from the standard. For this exercise, it's that the poverty rate is 12%.
  • Alternative Hypothesis ( \( H_a \) ): Believes the state's rate is different. Here, it鈥檚 less than 12%.
  • Data Collection: Gather a sample size. In this case, 1550 residents were examined for poverty conditions.
  • Calculation: Compute the sample proportion, then use statistical tests to see if it deviates from 12%.
This method is particularly useful for assessing changes or differences in a known population's behavior or characteristics.
Significance Level
The significance level, denoted as \( \alpha \) , is the threshold for deciding when to reject the null hypothesis. It's a measure of how willing one is to be wrong when saying there is a significant difference.The common choice for this level is 5%, meaning there's a 5% risk of concluding that a difference exists when there actually isn't one. When the p-value of the test is smaller than this preset alpha, the null hypothesis is rejected as unlikely.
  • 5% Significance Level: In the exercise, this means you'd accept a 5% chance of error in saying the state's poverty rate is below 12%.
  • Decision Making: It involves comparing the p-value to 0.05 to decide on the null hypothesis.
A lower significance level results in stronger conclusions, as there is a smaller chance of making a mistake.
P-value
The p-value is a statistical measurement to assess evidence against the null hypothesis. It gauges the probability of obtaining a result at least as extreme as the one observed, assuming the null hypothesis is true.
  • Low P-value: A small p-value indicates strong evidence against the null hypothesis, suggesting the sample's result is unlikely under the null hypothesis.
  • Calculation: In the exercise, a p-value of approximately 0.0257 was found for the z-value of -1.948. This suggests there is enough evidence to reject the null hypothesis.
  • Comparison: If the p-value is less than the significance level (in this case 0.05), the null hypothesis is rejected. Here, 0.0257 < 0.05, so it supports the alternative hypothesis.
The goal is to find a p-value which, if small, gives confidence to make a decision against the null hypothesis.
Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a default or baseline assumption about a population parameter. It is typically a statement of no effect or no difference. When conducting hypothesis testing, statisticians are testing this statement to see if evidence exists to dispute it.In this exercise:
  • Stated Hypothesis: The null hypothesis assumes the state's poverty rate equals the national rate of 12%, \( p = 0.12 \).
  • Purpose: It's a starting point for statistical testing. Initially, we assume it might be correct but look for evidence to prove otherwise.
  • Testing: Our results, a sample poverty rate of \( \hat{p} = 0.105 \) and subsequent calculations, provide statistical evidence to test \( H_0 \).
The goal is to analyze sample data to see if there is convincing reason to believe that the null hypothesis is false.

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Most popular questions from this chapter

In a recent year the fuel economy of all passenger vehicles was \(19.8 \mathrm{mpg}\). A trade organization sampled 50 passenger vehicles for fuel economy and obtained a sample mean of 20.1 mpg with standard deviation \(2.45 \mathrm{mpg}\). The sample mean 20.1 exceeds \(19.8,\) but perhaps the increase is only a result of sampling error. a. Perform the relevant test of hypotheses at the \(20 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(20 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

The recommended daily calorie intake for teenage girls is 2,200 calories/day. A nutritionist at a state university believes the average daily caloric intake of girls in that state to be lower. Test that hypothesis, at the \(5 \%\) level of significance, against the null hypothesis that the population average is 2,200 calories/day using the following sample data: \(n=36, x-2,150, s=203\).

Perform the indicated test of hypotheses, based on the information given. a. Test \(\mathrm{Ho}: \mu=212\) vs. Ha: \(\mu<212 @ \alpha=0.10, \sigma\) unknown, \(n=36, x-211.2, \mathrm{~s}=2.2\) b. Test \(H_{0}: \mu=-18\) vs. Ha: \(\mu>-18\) \(@ \alpha=0.05, \sigma=3.3, n=44, x-=-17.2, s=3.1\) c. Test \(H_{0}: \mu=24\) vs. Ha: \(\mu \neq 24 @ \alpha=0.02, \sigma\) unknown, \(n=50, x-=22.8, s=1.9\)

A special interest group asserts that \(90 \%\) of all smokers began smoking before age \(18 .\) In a sample of 850 smokers, 687 began smoking before age 18 . a. Test whether the true proportion of all smokers who began smoking before age 18 is less than \(90 \%,\) at the \(1 \%\) level of significance. b. Compute the observed significance of the test.

The target temperature for a hot beverage the moment it is dispensed from a vending machine is \(170^{\circ} \mathrm{F}\). A sample of ten randomly selected servings from a new machine undergoing a pre- shipment inspection gave mean temperature \(173^{\circ} \mathrm{F}\) with sample standard deviation \(6.3^{\circ} \mathrm{F}\). a. Assuming that temperature is normally distributed, perform the test that the mean temperature of dispensed beverages is different from \(170^{\circ} \mathrm{F}\), at the \(10 \%\) level of significance. b. The sample mean is greater than 170 , suggesting that the actual population mean is greater than \(170^{\circ} \mathrm{F}\). Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)
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