Scenario: Erna Economist claims that a majority of Americans feel that corporations make too muchprofit. According to a 2019 Pew Research poll, 56% of a random sample of 1,500 adults said thatbusiness corporations make too much profit. Run a hypothesis test to assess Erna Economist's claim.Question: What is the value of the test statistic, Z?

Given data sample size, n =1500 population proportion,p= 0.5 Significance level,…

Answered: nario: Erna Economist claims that a…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Classify the hypothesis test as two-tailed, left-tailed, or right-tailed. At one school, the average amount of time that tenth-graders spend watching television each week is 20.1 hours. The principal introduces a campaign to encourage the students to watch less television. One year later, the principal wants to perform a hypothesis test to determine whether the average amount of time spent watching television per week has decreased from the previous mean of 20.1 hours.
Burger King has recently added new menu items in the last few years. Q. How are you going to decide which menu items are potential winners? What statistical tests or inferential statistics can you conduct?
In 1970, 590 students among 1000 randomly selected college freshmen thought that capital punishment should be abolished. In 2005, 350 students among 1000 randomly selected college freshmen thought that capital punishment should be abolished. What is the two-sample z test statistic for evaluating the null hypothesis that the percentage of students who support capital punishment did not change from 1970 to 2005? Round your answer to two decimal places.
A political scientist claims that 38% of first-year college students characterize themselves as being “moderate” or “middle of the road” as far as their political affiliation is concerned. Believing this claimed value is too high, you survey a random sample of 400 first-year college students and find that 120 characterize themselves as being “moderate” or “middle of the road.” Based on this information, what will the test statistic be? Choose the answer below that is closest to what you calculate, and try not to do a lot of rounding until you get to the very end of your calculations. 1. -0.3 2. -1.2 3. -2.6 4. -3.3 5. None of the other answer options are correct because the test statistic should be positive, not negative.
In Advanced Statistical Inference: What is the relationship between the sufficiency, ancillary, and completeness?
6. Explain the reasoning behind announcing one's choice of using a one-tailed test before performing the statistical analysis on an observed data set.
According to the most recent General Social Survey, 70% of adults watch TV nightly. A researcher believes that the proportion of college students who watch TV nightly is less than this. To test this claim the research samples 400 college students and finds that 276 watch TV nightly. At the 0.05 level of significance does this data provide evidence to suggest that the proportion of college students who watch TV nightly is less than 70%? What is the p-value for this hypothesis test? Round to three decimal places.
According to a report on consumer fraud and identity theft, 20% of all complaints for a year were for identity theft. In that year, California had 724 complaints of identity theft out of 3466 consumer complaints. Does this data provide enough evidence to show that California had a higher proportion of identity theft than 20%? Test at the 5% level. State the hypotheses. Ho: P ? ✓ Ha: P?V Calculate the test statistic. Round to four decimal places. Ô = Calculate the standardized test statistic. Round to three decimal places. Z= Find the p-value. Round to four decimal places. p-value = State your decision. O Since the p-value is less than .05, reject Ho. O Since the p-value is less than .05, fail to reject Ho. Since the p-value is greater than .05, fail to reject Ho. Since the p-value is greater than .05, reject Ho. Interpret the results. O At the 5% level of significance, there is not enough evidence to show that the proportion of complaints due to identity theft in California is not…
A researcher believes that the method of instruction affects how well nursing students learn clinical skills. A total of N=20 nursing students are randomly assigned to learn clinical skills in one of four ways (texts, simulations, observations, or lectures), and their scores on a quiz on these clinical skills are assessed for each nursing student. The results are below. Use ?=.05 a. state the hypotheses (both in words and symbols) b. complete the ANOVA table above c. what should she conclude, use α = .05? (be sure to write the answer in APA format & Include F crit
According to a report on consumer fraud and identity theft, 20% of all complaints for a year were for identity theft. In that year, Rhode Island had 717 complaints of identity theft out of 3498 consumer complaints. Does this data provide enough evidence to show that Rhode Island had a higher proportion of identity theft than 20%? Test at the 5% level. State the hypotheses. Họ: p ? Ha: P Calculate the test statistic. Round to four decimal places. Calculate the standardized test statistic. Round to three decimal places. z = Find the p-value. Round to four decimal places. p-value = State your decision. O Since the p-value is less than .05, reject Ho. O Since the p-value is greater than .05, fail to reject Hg. O Since the p-value is less than .05, fail to reject Hg. O Since the p-value is greater than .05, reject Ho. Interpret the results. O At the 5% level of significance, there is enough evidence to show that the proportion of complaints due to identity theft in Rhode Island is more than…
A manager claims that fewer than 65% of employees will qualify for performance bonus this year. From a representative sample of 40 employees, 57.5% qualify for performance bonus. Test the manager’s claim at a 1% level of significance.

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