/*! 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 73 A number of initiatives on the t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A number of initiatives on the topic of legalized gambling have appeared on state ballots. Suppose that a political candidate has decided to support legalization of casino gambling if he is convinced that more than twothirds of U.S. adults approve of casino gambling. USA Today (June 17,1999 ) reported the results of a Gallup poll in which 1523 adults (selected at random from households with telephones) were asked whether they approved of casino gambling. The number in the sample who approved was 1035 . Does the sample provide convincing evidence that more than two-thirds approve?

Short Answer

Expert verified
After calculating the Z-score, if it's value is greater than 1.645, sufficient evidence exists to support the claim that more than two-thirds in favor of casino gambling. Otherwise, the evidence is not convincing enough.

Step by step solution

01

State the hypotheses

First, we state the null hypothesis (H0) and the alternative hypothesis (H1). In this case, H0 is \( P \leq 2/3 \) which means that two-thirds or less of all adults approve of casino gambling. The alternative hypothesis H1 is \( P > 2/3 \) which means that more than two-thirds of adults approve of casino gambling.
02

Calculate the test statistic

Next, we'll calculate the test statistic which is Z. The formula for Z is ( p - P0 ) / sqrt [ ( P0(1 - P0) ) / n ] where 'p' is the sample proportion, 'P0' is the population proportion (in this case two-thirds), and 'n' is the sample size. From our given data, p = 1035 / 1523 = 0.6797, P0 = 2/3 = 0.6667, and n = 1523. Plugging these into the formula will give us the calculated Z-score.
03

Determine the critical value

As the question wants to know if there is evidence supporting more than two-thirds, this is a one-sided test (right-tail). By using a Z-table or calculator, we find that the critical value for a 5% significance level is 1.645. This would be our rejection region. It means any Z value above 1.645 will fall into this region and we will reject our null hypothesis.
04

Compare the test statistic with the critical value

Comparing our calculated Z-value to our critical value will allow us to decide whether or not we reject the null hypothesis
05

Conclude

If the calculated Z value is greater than the critical value, we reject the null hypothesis and there's enough evidence to support that more than 2/3 approve. If the calculated Z value is less than or equal to the critical value, we fail to reject the null hypothesis and the support is not significant enough for the candidate to back casino gambling.

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 and Alternative Hypothesis
The null and alternative hypotheses are foundational components of hypothesis testing, a method used to determine if there is enough evidence in a sample of data to infer that a certain condition is true for the entire population.

The null hypothesis (\( H_0 \)), is a statement that there is no effect or no difference, and it represents a skeptical perspective or a claim to be tested. In our scenario, the null hypothesis is that two-thirds or less of U.S. adults approve of casino gambling, or \( P \leq \frac{2}{3} \).

The alternative hypothesis (\( H_1 \) or \( H_a \) ), is a statement that indicates the presence of an effect or a difference. It's what you aim to support. For the provided scenario, the alternative hypothesis tests whether more than two-thirds (\( P > \frac{2}{3} \) ) of the population approve of casino gambling.

Establishing these hypotheses is a crucial first step in the process of hypothesis testing, as they define the focus of the study and dictate the direction of the statistical analysis.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to decide whether to reject the null hypothesis.

In the context of our gambling poll example, the Z-score serves as the test statistic. It measures how many standard deviations an element is from the mean. The formula for computing the Z-score in a proportion test is: \[ Z = \frac{ p - P_0 }{ \text{sqrt} [ \frac{ P_0(1 - P_0) }{ n } ] } \], where 'p' is the sample proportion, \( P_0 \) is the assumed population proportion under the null hypothesis, and 'n' is the sample size.

The calculated Z-score is then compared to the critical value to determine whether the results are statistically significant—providing evidence to support or reject the null hypothesis.
Sample Proportion
The sample proportion is the percentage of individuals in the sample who exhibit a particular attribute of interest. It is denoted by \( p \) and is a key component in many statistical hypotheses tests, including the Z-test for proportions.

In our casino gambling example, the sample proportion is the number of adults who approve of casino gambling (\( 1035 \) adults) divided by the total number of adults in the sample (\( 1523 \) adults), so \( p = \frac{1035}{1523} \approx 0.6797 \). This sample proportion is what is utilized to calculate the test statistic, which is a crucial step in determining whether the null hypothesis can be rejected based on the data collected from the sample.
Critical Value
The critical value is a threshold that determines the boundary or cut-off point for rejecting the null hypothesis. It's derived from the desired level of significance (\( \alpha \) ), which is the probability of rejecting the null hypothesis when it is actually true (also known as a type I error).

In hypothesis testing, we compare the test statistic against the critical value. If the test statistic falls within the critical region (beyond the critical value), we reject the null hypothesis. For our one-sided test, the critical value at a 5% significance level is 1.645. This indicates that if the calculated Z-score is greater than 1.645, the evidence is in favor of the alternative hypothesis that more than two-thirds of U.S. adults approve of casino gambling.

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

A hot tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most 15 min. A random sample of 25 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(17.5 \mathrm{~min}\) and \(2.2 \mathrm{~min}, \mathrm{re}-\) spectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level \(.05 .\)

A certain television station has been providing live coverage of a particularly sensational criminal trial. The station's program director wishes to know whether more than half the potential viewers prefer a return to regular daytime programming. A survey of randomly selected viewers is conducted. Let \(\pi\) represent the true proportion of viewers who prefer regular daytime programming. What hypotheses should the program director test to answer the question of interest?

A well-designed and safe workplace can contribute greatly to increasing productivity. It is especially important that workers not be asked to perform tasks, such as lifting, that exceed their capabilities. The following data on maximum weight of lift (MWOL, in kilograms) for a frequency of 4 lifts per minute were reported in the article "The Effects of Speed, Frequency, and Load on Measured Hand Forces for a Floor-to-Knuckle Lifting Task" (Ergonomics \([1992]: 833-843):\) \(\begin{array}{lllll}25.8 & 36.6 & 26.3 & 21.8 & 27.2\end{array}\) Suppose that it is reasonable to regard the sample as a random sample from the population of healthy males, age \(18-30\). Do the data suggest that the population mean MWOL exceeds 25 ? Carry out a test of the relevant hypotheses using a \(.05\) significance level.

According to a Washington Post- \(A B C\) News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bothered if the National Security Agency collected records of personal telephone calls they had made. Is there sufficient evidence to conclude that a majority of U.S. adults feel this way? Test the appropriate hypotheses using a .01 significance level.

Although arsenic is known to be a poison, it also has some beneficial medicinal uses. In one study of the use of arsenic to treat acute promyelocytic leukemia (APL), a rare type of blood cell cancer, APL patients were given an arsenic compound as part of their treatment. Of those receiving arsenic, \(42 \%\) were in remission and showed no signs of leukemia in a subsequent examination (Washington Post, November 5,1998 ). It is known that \(15 \%\) of APL patients go into remission after the conventional treatment. Suppose that the study had included 100 randomly selected patients (the actual number in the study was much smaller). Is there sufficient evidence to conclude that the proportion in remission for the arsenic treatment is greater than .15, the remission proportion for the conventional treatment? Test the relevant hypotheses using a \(.01\) significance level.

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.