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- A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random sample of 1000 workers showed that 89% had paid leave. The resulting p-value is .0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012. What type of error is possible in this situation?Is there sufficient evidence to indicate the overall model is statistically useful for predicting y. Test using α=10%. Make sure you include, in your answer, the null and alternative hypotheses, the value of an appropriate test statistic, the p-value AND rejection region, and conclusion. Find the 95% confidence interval for β3. Use the interval to determine if mood scores would be useful to the model if both intrapersonal and stress management scores are already included in the model and give a brief explanation. TestH0: β2=0 against H0: β2>0 at α=05. Make sure you include, in your answer, the value of an appropriate test statistic, p-value AND rejection region, and conclusion.A publishing company has developed a study guide (reference book) for GRE Math test. They want to test if this reference book does help students perform better on the test. The company recruited a random sample of students who are planning to take this test and let them use the new reference book. Their scores are listed below. Test the null hypothesis that the students' mean score is not different from the national average (at a significance level of .05) by answering the following questions. The national statistics indicates that the average score for the population is 68.5. Scores: 70, 68, 93, 78, 73, 99, 81, 79, 61 (1) Set up hypotheses ( (2) What is the df and the critical value for this test? ( (3) What is the standard error for this test? (4) What is the t_observed for this test? ( (5) Provide the decision and the interpretation
- A mobile game currently has 38% of the market share. The CEO of the company producing the game recently authorized an advertising campaign with the goal of increasing the game's market share. A random sample of 1041 mobile gamers found that 409 gamers played the game at least once a day. A .Setup the null and alternative hypotheses to see if there is a significant change in the market share of the game. B. Determine if you can use a normal model to approximate the distribution of sample proportions. C. Find the critical value for a level of significance of 0.03. D. Find the standardized test statistic. E. State the conclusion of the hypothesis test.The U.S. Postal Service (USPS) is faced with the issue of a large portion of mail (letters and packages) that cannot be delivered because of an illegible or non-existent address. USPS would like to find out if there is any evidence that the proportion of undeliverable mail has significantly decreased since last year when it was 6%. Which of the following are the correct null and alternative hypotheses for this test? Select one: a. Ho: p = 0.06 versus Ha:p> 0.06 b. Ho: p 0.06 ge Next page Moodle Accessibility Specification Coastal Carolina University (CCU) does not discriminate on the basis of race, color, religion, sex, sexual orientation, genc national origin, age, genetic information, mental or physical disability, or status as a disabled or Vietnam-era veteran in it programs, activities or employment practices. For more information related to discrimination, please contact Title IX, Titl office phone 843-349-2382: FEN email eenarnastal adur or thel Dent of Education office for…What is the null hypothesis for a two-tailed test of Pearson's r correlation coefficient and how is it tested using a significance level of alpha=0.05?
- Indicate whether CPAs in states that have flat state income tax rates work fewer hours per week during tax season compared to the US average. You will want to formally set up H0 and Ha, report either the test statistic (z-value) or p-value, and explain why you either reject or fail to reject your null hypothesis.Kaplan Learning Center offers a GRE prep class. To gauge its effectiveness, Kaplan takes a random sample of 85 people enrolled in the prep class. The average GRE score for each student in the sample before enrolling in the prep class is compared to the average GRE score taken upon completion of the course. Which test of significance should be used to determine if there is a statistically significant difference? Why? Should it be one or two tailed?A new method was developed to reduce variability of test scores by eliminating lower scores. Two groups, one called the control group and the other, the experimental group, both took the same test. The experimental group was taught using the new method. Do the data provide sufficient evidence to conclude that there is less variation among scores when the new method is used? Perform an F-test at the 1% significance level. (Note: s, = 6.3 and Control Experimental 30 19 29 26 33 36 29 18 33 31 S2 = 3.3.) 33 18 19 32 27 32 29 30 33 ... .. First find the null and alternative hypotheses. Which of the following correctly states the hypotheses? O A. Ho: 01 72 Ha: 01 = 02 O B. Ho: 01 >02 Ha: 0102 Ha: 01 #02 O E. Ho: 01 =02 Ha: 01 02 nts SCO Compute the test statistic F. Scor (Round to three decimal places as needed.) ess
- An ANOVA provides the following data: Degrees of freedom between groups is 2; degrees of freedom within groups is 27; F- obt is 3.31 What is the correct statistical conclusion here? (hint: you'll need to find F-crit before you answer) O We fail to reject the null hypothesis and conclude that there is a significant difference among the groups' means O We reject the null hypothesis and conclude that there is a significant difference among the groups' means O We fail to reject the null hypothesis and conclude that there is no significant difference among the groups' means O We reject the null hypothesis and conclude that there is no significant difference among the groups' meansA new method of postoperative treatment was evaluated for patients undergoing a certain surgical procedure. Under the old method, the mean length of hospital stay was 6.3 days. The sample mean for the new method was 6.1 days. A hypothesis test was performed in which the null hypothesis stated that the mean length of stay for the new method was greater than or equal to the mean length for the old method, and the alternate hypothesis stated that the mean stay for the new method was lower. The P-value was 0.002. True or false: a) Because the P-value is very small, we can conclude that the new method provides an important reduction in the mean length of hospital stay. b) Because the P-value is very small, we can conclude that the mean length of hospital stay is less for patients treated by the new method than for patients treated by the old method.In 2018, a study was conducted that stated 22.6% of residents in Kern County live below the poverty line. Suppose you were hired to determine if that percentage has changed since the study was conducted. You randomly sample 200 residents from Kern County and find that 35 of them fall below the poverty line. Carry out the appropriate hypothesis test at the a = 0.05 level of significance to determine if the true proportion of Kern County residents that live below the poverty line is lower than what it was in 2018.