8. In tests of cheese production C is found that the mean time for a modification is made that is supp Tests on a random sample of 10 (in hours) for the production pro 522 492 541 502 At the 0.05 significance level, mean time for the production r
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- A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate j using the mean of the sample. What is the minimum sample size needed in order for the union to be 99% confdent that its estimate is within $0.40 of u? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.10. Carry your intermediate computations to at least three decimal places. Write your answer as a whole nuhber (and make sure that it is the minimum whole number that satisfies the requirements). (If necessary, consult a list of formulas.)A random sample of 100 private companies locating in Ankara was asked to rate on a scale from 1 (not important) to 5 (extremely important) for having the practical knowledge of statistics as a good job candidate. The sample mean rating was 4.7. a)Test at the 5% significance level the null hypothesis that the population mean rating is less than or equal to 4.5 against the alternative that it is greater than 4.5. Population variance is known as 0.6. b)Compute p-value and explain what the calculated p-value means.In tests of a computer component, it is found that the mean time between failures is 909 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 25 modified components produce a mean time between failures of 955 hours. Using a 1% level of significance, perform a hypothesis test to determine whether the mean time between failures for the modified components is greater than 909 hours. Assume that the population standard deviation is 53 hours.
- II-Term Exam Il-Make-up > الوقت المتبقي 40: The aim of a study is to test the ratio of variances in the weight of two groups of rats being under a certain drugs A (group 1) and B (group 2). A random sample of 17 rats was selected from group 1and 11 rats from group 2, Then their weight were recorded. The results are shown in the following table sd Median Mean n 1.05 3.5 17 group 1 2.0 2.20 11 group 2 3.0 1.15 F19,16,0.025 =0.386 F15,12,0.05=0.404 F16.10,0.025 = 0.335 F19,16,0.05=0.451 F15,12,0.1=0.496 | F16,10,0.05 = 0.401 F16,19,0.025 0.371 F12,15,0.05 =0.382 F10,16,0.025 =0.286| F16,19,0.05=0.437 F12,15,0.1=0.475 F10,16,0.05=0.354 Construct a 95% confidence interval for the ratio of the variances. [ Write only the final result in the box below ]The average retirement age in America is 62 years old. Do small business owners retire at a different average age? The data below shows the results of a survey of small business owners who have recently retired. Assume that the distribution of the population is normal. 69, 62, 73, 55, 74, 56, 60, 65, 75, 59, 66, 57, 61, 68 What can be concluded at the the αα = 0.01 level of significance level of significance? For this study, we should use Correct The null and alternative hypotheses would be: H0:H0: Correct Correct Correct H1:H1: Correct Correct Correct The test statistic Correct = Incorrect (please show your answer to 3 decimal places.) The p-value = Incorrect (Please show your answer to 4 decimal places.) The p-value is Correct αα Based on this, we should Correct the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton mean is significantly different from 62 at…Before the furniture store began its ad campaign, it averaged 158 customers per day. The manager is investigating if the average is smaller since the ad came out. The data for the 11 randomly selected days since the ad campaign began is shown below: 147, 142, 149, 133, 146, 159, 132, 137, 148, 172, 150 Assuming that the distribution is normal, what can be concluded at the αα = 0.10 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer > ≠ < = H1:H1: ? p μ Select an answer < ≠ > = The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? ≤ > αα Based on this, we should Select an answer accept fail to reject reject the null hypothesis. Thus, the final conclusion is that ... The data suggest…
- Facebook: A study showed that two years ago, the mean time spent per visit to Facebook was 20.8 minutes, Assume the standard devlation is o= 8.0 minutes. Suppose that a simple random sample of 107 visits was selected this year and has a sample mean of x= 18.8 minutes. A social scientist is interested to know whether the mean time of Facebook visits has decreased. Use the a=0.10 level of significance and the critical value method with the table. (a) State the appropriate null and alternate hypotheses. (b) Compute the value of the test statistic. (c) State a conclusion. Use the a = 0.10 level of significance.In an automobile repair and service industry, the average number of defective items in all the lots of automobile parts produced by a machine last week was 8. The supervisor believes that the mean number of defective parts produced by the same machine will differ in the present week. To test whether the average differs from 8, he samples 30 lots to obtain the mean number of defective parts to be 9 with a standard deviation of 3.Using the critical value approach, identify the decision at the 5% level of significance. Group of answer choices The decision is not to reject H0 because the test statistic does not fall in the rejection region (-∞,-1.960] U [1.960, ∞). The decision is to reject H0 because the test statistic falls in the rejection region (-∞,-1.645]. The decision is not to reject H0 because the test statistic does not fall in the rejection region [1.645, ∞). The decision is not to reject H0 because the test statistic does not fall in the rejection region (-∞,-1.645] U…15.34 (EX) Deer mice: Deer mice (Peromyscus maniculatus) are small rodents native to North America. Their body lengths (excluding tail) are known to vary approximately Normally with mean μ = 86 mm and standard deviation o = 8 mm. Deer mice are found in diverse habitats and exhibit different adaptations to their environment. A random sample of 14 deer mice in a rich forest habitat gives an average body length of x = 91.1 mm. Assume that the standard deviation o of all deer mice in this area is also 8 mm. In Exercise 14.35 (page 357) you gave a confidence interval based on the body lengths of 14 deer mice (Peromyscus maniculatus) from a rich forest habitat. Before you can trust your results, you would like more information about the data. What facts would you most like to know? Select all that apply. (Hint: There are 3 correct answers.) Whether body lengths of deer mice are exactly Normally distributed. Whether the sample of deer mice was random. We would like to verify that no…
- 9 -It is known that the products produced in a business are designed to have an average of 50gr and a variance of 16. The standard deviation of the randomly selected sample of 1000 for the quality control of the products from this enterprise was found to be 4 gr and the average 45.68 gr. The most common value is 45 gr, and when the data is put in order of magnitude, the middle value is 45.72 gr. The sample standard error value was found to be 0.126.What are the values of the sample statistics given in the question? a)4 - 45.68 - 45 - 45.72 B)1000 - 4 - 45.68 - 45 - 45.72 NS)50 - 16 D)4 - 45.68 - 45 - 45.72 - 0.126 TO)1000 - 45.68 - 45 - 45.72 - 0.1263. Overa long period of time, it was found that the time taken for the standard physical task is normally distributed with mean 50 seconds, and variance 100 sec². A sample of 75 workers were given a special and unusual diet for three weeks before they were observed performing said standard physical task. The mean time taken by this sample is 46 seconds. Is this significant evidence that the special diet has made a difference in their performance. Use a 5%level of significance.17.34 (EX) Oligofructose and calcium absorption 1/6: Nondigestible oligosaccharides are known to stimulate calcium absorption in rats. A double-blind, randomized experiment investigated whether the consumption of oligofructose similarly stimulates calcium absorption in healthy male adolescents 14 to 16 years old. The subjects took a pill for nine days and had their calcium absorption tested on the last day. The experiment was repeated three weeks later. Some subjects received the oligofructose pill in the first round and then a pill containing sucrose (which served as a control). The order was switched for the remaining subjects. Here are the fractional calcium absorption data (in percent of intake) for 11 subjects: Subject Control 1 2 3 4 5 6 7 8 9 10 11 78.4 76.6 57.4 51.5 49.0 46.6 44.2 42.9 37.2 34.1 24.6 Oligofructose 62.0 95.1 46.5 49.4 89.7 43.8 50.3 51.6 66.6 52.7 54.0 Explain clearly why the matched pairs t test is the proper choice for this experimental design. O Because this…