A furniture store claims that a specially ordered product will take, on average, u = 35 days (5 weeks) to arrive. The standard deviation of these waiting times is8 days. We suspect that the special orders are taking longer than this, To test this suspicion, we track a random sample of 60 special orders and find that theorders took a mean of 36 days to arrive. Can we conclude at the 0.10 level of significance that the mean waiting time on special orders at this furniture storeexceeds 35 days?Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)(a) State the null hypothesis H, and the alternative hypothesis H,.Ho(b) Determine the type of test statistic to use.(Choose one) ▼自ロO

From the given information : population mean, μ= 35.0 population standard deviation,σ= 8.0 sample…

Answered: A furniture store claims that a…

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 certain breed of rat shows a mean weight gain of 65g during the first 3 months of life. A random sample of 34 rats of this breed are fed a new diet from birth until age 3 months. These 34 rats have a mean weight gain of 60.75g and a standard deviation of 3.84g. We are interested in testing whether there is reason to believe, at the 0.05 significance level, that the new diet is causing a change in the average amount of weight gained in this breed of rats. By comparing the test statistic and the critical value, what conclusion can we draw at the 0.05 significance level? Since the test statistic is more extreme than the critical value, we reject the null hypothesis. We have statistically significant evidence to conclude that average weight gain during the first 3 months of a certain breed of rat's life on the new diet is not equal to 65g. Since the test statistic is NOT more extreme than the critical value, we fail to reject the null hypothesis. We do not have statistically significant…
The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.1 years and a standard deviation of 0.5 years. He then randomly selects records on 42 laptops sold in the past and finds that the mean replacement time is 3.9 years.Assuming that the laptop replacement times have a mean of 4.1 years and a standard deviation of 0.5 years, find the probability that 42 randomly selected laptops will have a mean replacement time of 3.9 years or less.P(M < 3.9 years)
The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 3.6 years and a standard deviation of 0.4 years. He then randomly selects records on 48 laptops sold in the past and finds that the mean replacement time is 3.4 years. Assuming that the laptop replacement times have a mean of 3.6 years and a standard deviation of 0.4 years, find the probability that 48 randomly selected laptops will have a mean replacement time of 3.4 years or less. P(M < 3.4 years)=__________
The average amount of time it takes for couples to further communicate with each other after their first date has ended is 3.31 days. Is this average shorter for blind dates? A researcher interviewed 63 couples who had recently been on blind dates and found that they averaged 3.2 days to communicate with each other after the date was over. Their standard deviation was 0.501 days. What can be concluded at the the α = 0.10 level of significance? For this study, we should use The null and alternative hypotheses would be: H0: H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is α Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly less than 3.31 at α = 0.10, so there is statistically insignificant evidence to conclude that the population mean time for couples who…
The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.1 years and a standard deviation of 0.6 years. He then randomly selects records on 41 laptops sold in the past and finds that the mean replacement time is 3.9 years.Assuming that the laptop replacement times have a mean of 4.1 years and a standard deviation of 0.6 years, find the probability that 41 randomly selected laptops will have a mean replacement time of 3.9 years or less.P(M < 3.9 years) = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.Based on the result above, does it appear that the computer store has been given laptops of lower than average quality?
The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 3.7 years and a standard deviation of 0.5 years. He then randomly selects records on 52 laptops sold in the past and finds that the mean replacement time is 3.6 years. Assuming that the laptop replacement times have a mean of 3.7 years and a standard deviation of 0.5 years, find the probability that 52 randomly selected laptops will have a mean replacement time of 3.6 years or less. P(M3.6 years) = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Based on the result above, does it appear that the computer store has been given laptops of lower than average quality? Yes. The probability of this data is unlikely to have occurred…
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The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.5 years and a standard deviation of 0.4 years. He then randomly selects records on 48 laptops sold in the past and finds that the mean replacement time is 4.4 years.Assuming that the laptop replacement times have a mean of 4.5 years and a standard deviation of 0.4 years, find the probability that 48 randomly selected laptops will have a mean replacement time of 4.4 years or less.P(M < 4.4 years) = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
An auto parts detailer uses a 5-week periodic review system to replenish its inventory stocks. The lead time is assumed to be 1 week since an order is placed to its receipt. The demand for an auto part during the 6 weeks (review period and lead time) is observed to be normally distributed with and mean of 200 units with a standard deviation of 5 units. If the manager places an order of 30 units with no other stocks for this product in-hand, what is provability that there will be stock out prior to the arrival of the replenishment? Round to 3-decimal places O a. None is correct O b. 0.023 O c. 0.421 O d. 0.500 O e. 0.726
Assume that an internet provider monitors the amount of time that every customer accesses their home internet connection each day. The amount of time that all of their customers access their home internet connection is normally distributed. They access it for an average of 8.9 hours per day with a standard deviation of 2.1 hours. What proportion of samples of 100 will spend less than an average of 9 hours on their home internet per day? Your answer should have 4 decimal places.
Are juvenile defendants who are released pending adjudication processed more slowly than those who are detained? The JDCC data set records whether a juvenile was released and the number of months it took for that juvenile’s case to reach a disposition. The sample will be narrowed to black female youth. Released juveniles (Sample 1) had a mean of 4.30 months to disposition (s = 3.86, N = 123) and detained juveniles (Sample 2) had a mean of 3.57 (s = 4.20, N = 64). Using an alpha level of .01, test the hypothesis that released juveniles’ mean time-to-adjudication is significantly greater than detained juveniles’ mean. Assume equal population variances. Use all five steps.
An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. How large asample is needed if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean?

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