/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 In January \(2016,\) at the end ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 (0.544,0.596). The z-score for the null hypothesis test is \(3.85\), which is greater than the critical value of \(1.96\), thus we reject the null hypothesis, confirming that Obama's approval rating has significantly changed from \(52\%\).

Step by step solution

01

Calculate standard error

First, the standard error is calculated using the formula: \(SE = \sqrt{p(1-p)/n}\),\n where p is the sample proportion and n is the sample size. Here, \(p=0.57\) and \(n=1500\), so the standard error is \(SE = \sqrt{0.57(1-0.57)/1500} = 0.013\).
02

Find confidence interval

Next, a 95% confidence interval is constructed for the approval rating. \nThe confidence interval is given by \(CI = p \pm z*SE\), where \(z = 1.96\) for a 95% confidence interval. Hence, the CI is \(CI = 0.57 \pm 1.96 * 0.013 = (0.544,0.596)\).
03

Null Hypothesis Testing

The null hypothesis stated that Obama's approval rating was essentially the same as his approval rating when he was elected to his second term, i.e., \(52\%\).\n A Z-test can be done to test this hypothesis.\n Z = (p-p0)/SE, where p0 = 0.52 (the null hypothesis). Hence, Z = (0.57-0.52)/(0.013) = 3.85. Comparing this value with the critical value of \(1.96\) (for α = 0.05) we could say that the Z value is farther from zero than the critical value. That means we reject the null hypothesis.
04

Interpretation

Based on the results, the 95% confidence interval for Obama's approval rating was between \(54.4\%\) and \(59.6\%\). Since the confidence interval does not include the hypothesized value of \(52\%\), there is enough evidence to reject the null hypothesis. So, Obama's approval rating was significantly different from \(52\%\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

null hypothesis testing
Null hypothesis testing is a fundamental concept in statistical analysis used to determine whether there's enough evidence in a sample of data to infer that a certain condition is true for the entire population. In the exercise, President Obama's approval rating serves as a real-world example where we can apply this concept.

When testing a null hypothesis, you are essentially testing the assumption that there is no effect or no difference - in our case, that Obama's approval rating at the end of his term is the same as when he was elected to his second term. To test this hypothesis, one must use a statistical test - here, it is the Z-test - to determine if the observed data, with a calculated test statistic, allows us to reject the null hypothesis or fail to reject it.

To translate the math into English: if the test result (the Z-value) is extreme enough, it suggests that the sample provides enough evidence to say there is a statistically significant difference in the approval ratings, prompting statisticians to reject the null hypothesis. When we reject the null hypothesis, we are saying that our sample provides strong evidence that the real approval rating is different from what was assumed under the null hypothesis.
standard error calculation
The standard error is a measure of the amount of variability or dispersion in a set of sample statistics. Calculating the standard error is vital as it gives us a sense of how accurately our sample estimates the true population parameter.

In our example, we are dealing with Obama's approval rating. To calculate the standard error (SE), you use the formula for a proportion: \( SE = \sqrt{\frac{p(1-p)}{n}} \) where \( p \) is the sample proportion (approval rating) and \( n \) is the sample size (number of surveyed individuals). Then by plugging in Obama's approval rating of \(57\%\) and the sample size of 1500, you get a standard error which helps in building the confidence interval.

The standard error gives us a grasp of how much we would expect Obama's approval rating to vary from one sample of 1500 U.S. adults to another. It's essential for standard error to be small, which would indicate our sample is highly likely to be representative of the whole population's approval rating for Obama.
Z-test statistics
Z-test statistics are used when we wish to compare the sample data to a population mean, assuming we know the population variance or when we have a large sample size. The Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean.

In President Obama's approval rating scenario, we computed a Z-score to see how much the sample proportion deviates from the null hypothesis proportion. Using the formula \( Z = \frac{p - p_0}{SE} \) where \( p \) is the sample proportion, \( p_0 \) is the null hypothesis proportion (approval rating when elected to the second term), and SE is the standard error, we arrive at a Z-score which we then compare to critical values from the standard normal distribution.

If the absolute value of our calculated Z-score is larger than the critical value, as it is in this case (3.85 as compared to 1.96), this suggests that the sample proportion is significantly different from the hypothesized population proportion. It's a way to quantify the 'unlikeliness' that the null hypothesis is true, guiding us to reject it and conclude a significant difference in approval ratings.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Testing for Alzheimer's disease can be a long and expensive process, consisting of lengthy tests and medical diagnosis. A group of researchers (Solomon et al., 1998 ) devised a 7-minute test to serve as a quick screen for the disease for use in the general population of senior citizens. A patient who tested positive would then go through the more expensive battery of tests and medical diagnosis. The authors reported a false-positive rate of \(4 \%\) and a false-negative rate of \(8 \%\). a. Put this in the context of a hypothesis test. What are the null and alternative hypotheses? b. What would a Type I error mean? c. What would a Type II error mean? d. Which is worse here, a Type I or Type II error? Explain. e. What is the power of this test?

A survey of 81 randomly selected people standing in line to enter a football game found that 73 of them were home team fans. a. Explain why we cannot use this information to construct a confidence interval for the proportion of all people at the game who are fans of the home team. b. Would a bootstrap confidence interval be a good idea?

Canine hip dysplasia is a degenerative disease that causes pain in many dogs. Sometimes advanced warning signs appear in puppies as young as 6 months. A veterinarian checked 42 puppies whose owners brought them to a vaccination clinic, and she found 5 with early hip dysplasia. She considers this group to be a random sample of all puppies. a. Explain why we cannot use this information to construct a confidence interval for the rate of occurrence of early hip dysplasia among all 6 -month- old puppies. b. Could you use a bootstrap hypothesis test? Why or why not?

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 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?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.