03.5479Determine the appropriate conclusion for this hypothesis test.O The sample data do not provide sufficient evidence to reject the alternative hypothesis that theaverage monthly electricity bill from this utility company is greater than $95.98 and thus weconclude that the average monthly electricity bill from this utility company is likely greater than$95.98.O The sample data do not provide sufficient evidence to reject the null hypothesis that the averagemonthly electricity bill from this utility company is $95.98 and thus we conclude that averagemonthly electricity bill from this utility company is likely about $95.98.O The sample data provide sufficient evidence to reject the alternative hypothesis that the averagemonthly electricity bill from this utility company is greater than $95.98 and thus we conclude thatthe average monthly electricity bill from this utility company is likely about $95.98, as claimed.O The sample data provide sufficient evidence to reject the null hypothesis that the average monthlyelectricity bill from this utility company is $95.98 and thus we conclude that average monthlyelectricity bill from this utility company is likely greater than $95.98.> Next Question A local utility company claims that the monthly electricity bills of its customers are among the lowest inthe state with an average monthly electricity bill of $95.98.An energy analyst believes that the average monthly electricity bill from this utility company is actuallyhigher than S95.98. The analyst collected a random sample of 39 monthly electricity bills from this utilitycompany and found that the average monthly bill of the sample was $105.58, with a standard deviation of$16.90.Use the critical-value method to test the hypothesis that the average monthly electricity bill for customersof this local utility is greater than $95.98, using a = 0.1.State the null and alternative hypothesis for this test.O Ho: µ > 95.98, H1:p= 95.98O Ho:µ = 95.98, H1:u > 95.98O Ho: µ = 95.98, H1:u 95.98O Ho: µ = 95.98, H1: u < 95.98Determine the critical-value for the hypothesis test.O 3.5657O 1.6859O 2.0244O1.3042Determine the test statistic for the hypothesis test.O 2.7424O 0.5681O 3.9324O 3.5479

Given data: Significance level = 0.1 Claim: An energy analyst believes that the average monthly…

Answered: A local utility company claims that the…

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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A local church parish wants to raise money to add to their campus. In a sample from a previous fund raising campaign, the parish found that of the 139 people in the sample they contacted, that 68 actually contributed money. Of those that contributed money, the average contribution was $898 with a standard deviation of $268. In the recent fundraising campaign, a sample of 90 people revealed that 52 have contributed money with an average contribution of $1,118 and a standard deviation of $354. When testing (at the 5% level of significance) whether the average amount contributed has increased in the current campaign compared to the previous campaign, what is the test statistic? (please round your answer to 2 decimal places)
Assume a media agency reports that it takes television streaming service subscribers in the U.S. an average of 6.14 days to watch the entire first season of a television series, with a standard deviation of 4.03 days. Scarlett is an analyst for an online television and movie streaming service company that targets to the 18-50 age bracket. She wants to determine if her company's customers exhibit shorter viewing rates for their series offerings. Scarlett plans to conduct a one-sample z-test, with significance level of a = 0.05, to test the null hypothesis, Ho: u = 6.14 days, against the alternative hypothesis, Hj: µ < 6.14 days. The variable u is the mean amount of time, in days, that it takes for customers to watch the first season of a television series. Scarlett selects a simple random sample of 825 customers who watched the first season of a television series from the company database of over 25,000 customers that qualified. She compiles the summary statistics shown in the table.…
According to a government agency, the average workweek for an adult in February 2018 was 34.1 hours. Assume the population standard deviation for the number of hours worked per week is 4.0 hours. A random sample of 35 adults worked an average of 34.8 hours last week.
Assume that human body temperatures are normally distributed with a mean of 98.23°F and standard deviation of .64°F. The hospital uses 100.6°F as the lowest temperature considered to be a fever. What percentage of normal and healthy person's would be considered to have a fever? Does this percentage suggest that a cut off of 100.6°F is appropriate? Physicians want to select a minimum temperature for requiring further medical test. What should the temperature be if we want only 5% of healthy people to exceed it?
A half-century ago, the mean height of women in a particular country in their 20s was 62.4 inches. Assume that the heights of today's women in their 20s are approximately normally distributed with a standard deviation of 1.67 inches. If the mean height today is the same as that of a half-century ago, what percentage of all samples of 27 of today's women in their 20s have mean heights of at least 63.02 inches? About% of all samples have mean heights of at least 63.02 inches. (Round to one decimal place as needed.)
The heights of 10 year old American children are normally distributed with a mean height of 4 feet and 5 inches (or 53 inches) and a standard deviation of 4.5 inches.In order to be allowed to go on a particular kids ride at the amusement park, children must be at least 44.9 inches tall and no more than 59.75 inches tall.What percent of 10 year old American children cannot go on the ride?
A local church parish wants to raise money to add to their campus. In a sample from a previous fund raising campaign, the parish found that of the 131 people in the sample they contacted, that 69 actually contributed money. Of those that contributed money, the average contribution was $857 with a standard deviation of $265. In the recent fundraising campaign, a sample of 95 people revealed that 48 have contributed money with an average contribution of $1109 and a standard deviation of $339. When testing (at the 5% level of significance) whether the average amount contributed has increased in the current campaign compared to the previous campaign, what is the test statistic? (please round your answer to 2 decimal places)
3) A SRS of 25 of the same model of car results in an annual mean maintenance cost of $575 and a standard deviation of $58. A car sales person claims the annual mean maintenance cost of that model is less than $600. Test the claim with a = .01.
In a certain city, the average 20- to 29-year old man is 72.5 inches tall, with a standard deviation of 3.2 inches, while the average 20- to 29-year old woman is 64.3 inches tall, with a standard deviation of 3.9 inches. Who is relatively taller, a 75-inch man or a 70-inch woman?
1) A half-century ago, the mean height of women in a particular country in their 20s was 63.7 inches. Assume that the heights of today's women in their 20s are approximately normally distributed with a standard deviation of 2.31 inches. If the mean height today is the same as that of a half-century ago, what percentage of all samples of 26 of today's women in their 20s have mean heights of at least 65.04 inches?
Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On eight different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 19.7 hours with a standard deviation of 9.4 hours. He measured the length of time his wife's cell phone battery lasted on twelve different occasions and calculated a mean of 18.9 hours with a standard deviation of 5.3 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Step 1 of 2: Construct a 90 % confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.

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