No prior information is needed (b) Now suppose that they have collected a second sample to test the same null hypothesis.Working with the data, they find that:ob 31PerssabihenplaZ.The test statistics of the new sample mean is:Plugging this test statistics into Excel yields:121Based on this new sample, circle all and only correct statement(s) below.No explanation is needed.(A) The null hypothesis can be rejected at the 68% confidence level.(B)) The null hypothesis can be rejected.(C) The p-value of the test is between 0.1 and 0.2.(D)The p-value of the test is between 0.25 and 0.4.(E) The p-value of the test is above 0.55.(F)The null hypothesis can be rejected when the significance level is 0.15.(G)The null hypothesis can be rejected when the significance level is 0.25.(H)The null hypothesis can be rejected at the 70% confidence level(1)The null hypothesis can be rejected at the 80% confidence level.(J)The null hypothesis can be rejected at the 90% confidence level.UPPERT1.12norm.s.dist(1.12, true)=0.87P-value1.12 20.87Reject4CVval:✓(0.84).

It is given that the test statistic is 1.12 and the corresponding P-value is 0.87.

Answered: (b) Now suppose that they have…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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For a two-tailed hypothesis test at the significance level α,the null hypothesis H0: μ=μ0 will be rejected in favor of the alternative hypothesis Ha: μ≠μ0 if and only if μ0 lies outside the (1−α)-level confidence interval for μ. Illustrate the preceding relationship by obtaining the appropriate one-mean t-interval for the summary statistics provided. Complete parts (a) and (b). According to a survey, the average person watched 4.35 hours of television per day in 2005. A random sample of 20 people were asked for the number of hours of television watched per day for last year. At the 10% significance level, the data do not provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from that in 2005. (Note: x=4.752 hours and= s=2.384 hours.) The confidence interval is (Round to three decimal places as needed.) According to a national study the average consumer unit spent $1739 on apparel and services…
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In exploring possible sites for a convenience store in a large neighbourhood, the retail chain wants to know the proportion of ratepayers in favour of the proposal. If the estimate is required to be within 0.1 of the true proportion, would a random sample of size n = 100 from the council records be sufficient for a 95% confidence interval of this precision?
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Which of these cells would correctly find a bootstrapped 95% confidence interval for the mean weight of a population? a. orig_sample = Table().read_table("Sample.csv") new_means = make_array() for i in np.arange(1000): new_mean = np.average(orig_sample.sample().column("Weight")) new_means = np.append(new_means, new_mean) Lower_bd = percentile(2.5, new_means)Upper_bd = percentile(97.5, new_means) print(f"The 95% confidence interval is from {Lower_bd} to {Upper_bd}") b. orig_sample = Table().read_table("Sample.csv") new_means = make_array() for i in np.arange(1000): new_mean = np.average(orig_sample.sample().select(Weight)) new_means = np.append(new_means, new_mean) Lower_bd = percentile(5, new_means)Upper_bd = percentile(95, new_means) print(f"The 95% confidence interval is from {Lower_bd} to {Upper_bd}") c. orig_sample = Table().read_table("Sample.csv") new_means = make_array() for i in np.arange(1000): new_mean = np.average(orig_sample.column(Weight))…
A researcher is interested in estimating the proportion of voters who favor a tax on e-commerce. ) ___________. Based on a sample of 250 people, she obtains the following 99% confidence interval for the population proportion, p: 0.113 < p < 0.171. Which of the statements below is a valid interpretation of this confidence interval? There is a 99% chance that the true value of implies between 0.113 and 0.171. If 100 different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, exactly 99 of these confidence intervals would contain the true value of p. If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, in the long run 99% of the confidence intervals would contain the true value of p. If many different samples of size 250 were selected and, based on each sample, a confidence intervaL were constructed, 99% of the time the true value of p would lie…
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