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Chapter 19: Problem 19

In January \(2016,\) at the end of his time in office, President Obama's approval rating stood at \(57 \%\) in Gallup's daily tracking poll of 1500 randomly surveyed U.S. adults. (www.gallup.com/poll/113980/gallup-daily-obama- jobapproval.aspx) a. Make a \(95 \%\) confidence interval for his approval rating by all U.S. adults. b. Based on the confidence interval, test the null hypothesis that Obama's approval rating was essentially the same as his approval rating of \(52 \%\) when he was elected to his second term.

Short Answer

Expert verified
The 95% confidence interval for Obama's approval rating is [0.5448, 0.5952]. After conducting a hypothesis test, we reject the null hypothesis that Obama's approval rating was 52%, which indicates that his approval rating was significantly higher.

Step by step solution

01

Calculate Standard Error

The Standard Error (SE) can be calculated using the formula: \(SE = \sqrt{\frac{{p*(1-p)}}{n}}\) where, p is the proportion (approval rating) and n is the sample size. Substituting given values into the formula: \(SE = \sqrt{\frac{{0.57*0.43}}{1500}} = 0.01293\).
02

Construct the confidence interval

The \(95\%\) confidence interval formula is given by \(p \pm z*SE\), where z for a \(95\%\) confidence level is \(1.96\). Substituting the values into the formula to calculate the confidence interval: \(0.57 \pm 1.96*0.01293\). This gives us the confidence interval as \([0.5448, 0.5952]\)
03

Hypothesis test

We will conduct a hypothesis test to check if Obama's approval rating was the same as his earlier rating of 52%. The null hypothesis is that the approval rating is 52%, i.e., \(p=0.52\). The test statistic (z) can be calculated using the formula: \(z = \frac{{p - p_0}}{SE}\). Substituting in the known values: \(z = \frac{{0.57 - 0.52}}{0.01293} = 3.87\). Comparing this to the critical z-value for a 95% confidence level (1.96), we reject the null hypothesis, because the test statistic is greater than the critical value. This means that Obama's approval rating was significantly different from 52% at the 95% confidence level.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. It involves making an initial assumption, called the null hypothesis, and then determining whether the observed data is significantly different from what the null hypothesis would predict. In the context of approval ratings, we might test the null hypothesis that a president鈥檚 current approval rating is equal to a previous rating, as was done with President Obama's approval ratings.
Standard Error
The standard error (SE) is a statistical term that measures the amount of variability or dispersion of a sample mean in relation to the true population mean. It is crucial in constructing confidence intervals and conducting hypothesis tests. The SE is derived from the standard deviation of the sample and the sample size, indicating how precise our estimate of the population parameter is. A smaller SE implies a more reliable estimate. In the exercise, the standard error is calculated from President Obama's approval rating and the sample size, representing how much we expect the sample proportion to deviate from the true population proportion.
Z-test
A z-test is a statistical test used to determine whether there is a significant difference between sample and population means, assuming that the sample is normally distributed, and the population standard deviation is known. When calculating a z-test statistic, we measure how many standard errors away from the population mean the sample mean falls. In our exercise, a z-test helps us to test whether President Obama's final approval rating significantly differed from the previous one by comparing the z-test statistic to the critical z-value associated with our chosen confidence level.
Gallup Poll
The Gallup Poll is a reputable survey organization that collects data on a variety of topics, including presidential approval ratings. These polls use random sampling of U.S. adults to estimate the opinions and behaviors of the entire population. It鈥檚 important to understand the methodology behind such polls to interpret their results correctly. The survey mentioned in the exercise provides the data point for President Obama's approval rating, which serves as the basis for constructing a confidence interval and performing hypothesis testing.
Approval Rating
Approval rating is a measure of how many people approve of a person鈥檚 performance in a specific role, commonly used to describe how the public feels about the performance of elected officials such as the President. It is typically expressed as a percentage, representing the proportion of respondents in a poll who approve of the person鈥檚 performance. In our exercise, we worked with President Obama鈥檚 approval rating as estimated by the Gallup Poll.
Null Hypothesis
The null hypothesis is the default assumption that there is no effect or no difference, which we seek to test against the alternative hypothesis that there is an effect or difference. In hypothesis testing, we attempt to determine whether the observed data is inconsistent with the null hypothesis to an extent that it can be rejected. In the context of approval ratings, the null hypothesis could assert that the current approval rating is not different from a known value, such as a past approval rating or a specific benchmark.

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Most popular questions from this chapter

A statistics professor has observed that for several years students score an average of 105 points out of 150 on the semester exam. A salesman suggests that he try a statistics software package that gets students more involved with computers, predicting that it will increase students' scores. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly. The professor will have to pay for the software only if he chooses to continue using it. a. Is this a one-tailed or two-tailed test? Explain. b. Write the null and alternative hypotheses. c. In this context, explain what would happen if the professor makes a Type I error. d. In this context, explain what would happen if the professor makes a Type II error. e. What is meant by the power of this test?

Which of the following are true? If false, explain briefly. a. If the null hypothesis is true, you'll get a high P-value. b. If the null hypothesis is true, a P-value of 0.01 will occur about \(1 \%\) of the time. c. A P-value of 0.90 means that the null hypothesis has a good chance of being true. d. AP-value of 0.90 is strong evidence that the null hypothesis is true.

An artist experimenting with clay to create pottery with a special texture has been experiencing difficulty with these special pieces. About \(40 \%\) break in the kiln during firing. Hoping to solve this problem, she buys some more expensive clay from another supplier. She plans to make and fire 10 pieces and will decide to use the new clay if at most one of them breaks. a. Suppose the new, expensive clay really is no better than her usual clay. What's the probability that this test convinces her to use it anyway? (Hint: Use a Binomial model.) b. If she decides to switch to the new clay and it is no better, what kind of error did she commit? c. If the new clay really can reduce breakage to only \(20 \%,\) what's the probability that her test will not detect the improvement? d. How can she improve the power of her test? Offer at least two suggestions.

A researcher developing scanners to search for hidden weapons at airports has concluded that a new device is significantly better than the current scanner. He made this decision based on a test using \(\alpha=0.05 .\) Would he have made the same decision at \(\alpha=0.10 ?\) How about \(\alpha=0.01 ?\) Explain.

A company is willing to renew its advertising contract with a local radio station only if the station can prove that more than \(20 \%\) of the residents of the city have heard the ad and recognize the company's product. The radio station conducts a random phone survey of 400 people. a. What are the hypotheses? b. The station plans to conduct this test using a \(10 \%\) level of significance, but the company wants the significance level lowered to \(5 \%\). Why? c. What is meant by the power of this test? d. For which level of significance will the power of this test be higher? Why? e. They finally agree to use \(\alpha=0.05,\) but the company proposes that the station call 600 people instead of the 400 initially proposed. Will that make the risk of Type II error higher or lower? Explain.
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