The St. Lucie County Economic Development Board needs to estimate…

Question Answered step-by-step The St. Lucie County Economic Development Board needs to estimate… The St. Lucie County Economic Development Board needs to estimate the proportion of all county residents who have earned at least a bachelor’s degree. A random sample conducted by the St. Lucie county Economic ∂development Board found that 20.9% of all county residents held at least a bachelor’s degree.Find three different confidence intervals- one with sample size 344, now with sample size 519, and one with sample size 757. Assume that 20.9% of the county residents in each sample have earned at least a bachelor’s degree. Notice how the sample size affects the margin of error and the width of the interval.Report confidence interval solutions using interval notation. Report all solution in percent form, round to two decimal places, if necessary.- When n = 344, the margin of error for a 98% confidence interval is given by ——- =When n = 344, a 98% confidence interval is given by —— =- When n = 519, the margin. of error for a 98% confidence interval is given by ——–=When n = 519, a 98%. confidence interval is given by —–=- When n = 757, the margin of error for a 98% confidence interval is given by —— =When n = 757, a 98% confidence interval is given by —– = 2- Calculate the critical z value (s) for each of the given hypothesis test scenarios below. If multiple critical values exist for a single scenario, enter the solutions using a comma-separated list. Round z-values to two decimal places. – Find the critical z-value for a left-tailed of hypothesis for a mean, assuming the population standard deviation is known, with a sample size of 127 and let alpha = 0.01.z =- Find the critical z-value (s) for a right-tailed test of hypothesis for a proportion, with a sample size of 91 and a significance level of 2.5%z = -Find the the critical z-value (s) for a two-tailed test of hypothesis for a mean, assuming the population standard deviation is know, with a sample size of 131 and a significance level of 10%z = – Find the critical value (s) for a two-tailed test hypothesis for a proportion, with a sample size of 121 and a significance level of 2%.z = 3- An LED light bulb manufacturer claims that the average lifespan of its 60-watt LED bulbs is 34,000 hours. A corporate watchdog group is suspicious of the company’s claim and thinks that the true average lifespan of the 60-watt LED bulbs produced by the company may be less than the advertised 34,000 hours. the group collected and tested a random sample of 291 light bulbs and found the average lifespan of the sample was 33,945 hours.Use the p-value method to test the hypothesis that the mean lifespan ofnthis brand of 60-watt LED light bulbs is less than 34,000 hours, using alpha = 0.05. Assuming the standard deviation of the lifespan of all such light bulbs is known to be 440 hours.State the null and alternative hypothesis for this test. Ho:? –H1:? –determine the test statistic for the hypothesis test. Round the solution to two decimal places =—–Determine the p-value for the hypothesis test. Round the solution to four decimal places.=—–Determine the appropriate conclusion for this hypothesis test.- The sample data do not provide sufficient evidence to reject the alternative hypothesis that the mean lifespan of the 60-watt LED light bulbs produced by this company is less than 34,000 hours and thus we conclude that the company’s claim that the light bulbs last an average of 34,000 hours is likely false.- The sample data do not provide sufficient evidence ti reject the null hypothesis that the mean lifespan of the 60-watt LED light bulbs produced by this company is 34,000 hours and thus we have conclude that the company’s claim that the average lifespan of the 60-watt LED light bulbs is 34,000 hours is likely true.- The sample data provide sufficient evidence to reject the null hypothesis that the mean lifespan of the 60-watt LED light bulbs produced by this company is 34,000 hours and thus we conclude that the company’s claim that the average lifespan of the 60-watt LED light bulbs is 34,000 hours is likely false 4- Hospital emergency rooms (ERs) have a bad reputation for a long wait times. To improve their image, many hospitals have worked hard to reduce they ER wait times and frequently advertise their improved, low ER wait times. Lawnwood regional Medical center recently advertised a 27 minutes or less wait times at their ER. An accreditation agency is suspicious of this low advertised wait time and would like to test the hospital’s claim.A random sample of 127 ER visit to Lawnwood Regional Medical Center was examined and the mean wait time of the sample was found to be 27.85 minutes.Using a significance level of 2%, can the accreditation agency conclude that the hospital’s claim about ER wait times is false? Assuming that the standard deviation of the wait times of all ER visits at Lawnwood Regional Medical Center is known to be 2.99 minutes. Use the critical value method.State the null and alternative hypothesis for this test. Ho:? —-H1:? —-Determine if this test islet-tailed, right-tailed, or two-tailed.- Determine the critical value (s) for the hypothesis test. Round the solution(s) to two decimal places. If more than one critical value exists, enter the solutions using a comma-separated list.=—-Determine the test statistic. Round the solution to two decimal places. =—— 5- According to a recent report published by the USDA 3 years ago, a typical American consumes an average of 29 pounds of French fries per year.A researcher at the USDA would like to determine if the average amount of French fries consumed each year by a typical American has changed since the original report was published. The researcher collected data from a random sample of 57 Americans and found that the mean amount of French fries consumed each year by the sample was 28.18 pounds with a standard deviation of 1.64.Using a significance level of 5%, test the hypothesis that the mean amount of French fries consumed each year by a typical American is different than 29 pounds. Use the p-value method.State the null and alternative hypothesis for this test.Ho:? —–H1:? —–Determine the test statistic for the hypothesis test. Round the solution to four decimal places.=—-Determine the p-value (range) for the hypothesis test.-p-value < 0.001-0.001 < p-value < 0.01-0.01 < p-value < 0.02-0.02 < p-value < 0.05-0.05 < p-value < 0.10-0.10 < p-value < 0.20-p-value < 0.20Determine the appropriate conclusion for this hypothesis test.The sample data provide sufficient evidence to reject the null hypothesis that the mean amount of French fries consumed each year by a typical American is 29 pounds and thus we conclude that the mean amount of French fries consumed each year by a typical American is likely different than 29 pounds.-nThe sample data do not provide sufficient evidence to reject the null hypothesis that the mean amount of French fries consumed by a typical American is 29 pounds and thus we conclude that the mean amount of French fries consumed each year by a typical American is likely 29 pounds.-The sample data do not provide sufficient evidence to reject the alternative hypothesis that the mean amount of French fries consumed by a typical American is different than 29 pounds and thus we conclude that the mean amount of French fries consumed each year by the typical American is likely different than 29 pounds. Math Statistics and Probability STA 2023 Share QuestionEmailCopy link Comments (0)