Heightsmeasured for a random sample of 10 plants grown while being treated with a particular nutrient. The sample mean and sample standard deviatiof those height measurements were 45 centimeters and 12 centimeters, respectively.Assume that the population of heights of treated plants is normally distributed with mean u. Based on the sample, can it be concluded that u is different from42 centimeters? Use the 0.05 level of significance.Perform a two-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list offormulas.)(a) State the null hypothesis H, and the alternative hypothesis H,.H, :0H :0(b) Determine the type of test statistic to use.(Choose one) ▼O=0OSO(c) Find the value of the test statistic. (Round to three or more decimal places.)(d) Find the two critical values. (Round to three or more decimal places.)I and(e) At the 0.05 level of significance, can it be concluded that the populationmean height of treated plants is different from 42 centimeters?

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|(a) State the null hypothesis H, and the alternative hypothesis H7. H, :0 H :0 |(b) Determine the type of test statistic to use. (Choose one) |(c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the two critical values. (Round to three or more decimal places.) D and 0

|(a) State the null hypothesis H, and the alternative hypothesis H7. H, :0 H :0 |(b) Determine the type of test statistic to use. (Choose one) |(c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the two critical values. (Round to three or more decimal places.) D and 0

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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State whether the standardized test statistic z indicates that you should reject the null hypothesis. (a) z= 1.225 (b) z = - 1.329 (c) z = - 1.517 (d) z = - 1.158 2 =-1285 (a) For z= 1.225, should you reject or fail to reject the null hypothesis? O A. Fail to reject H, because z - 1.285. O C. Fail to reject H, because z> - 1.285. O D. Reject H, because z<- 1.285.
The owner of a chain of mini-markets wants to compare the sales performance of two of her stores, Store 1 and Store 2. Sales can vary considerably depending on the day of the week and the season of the year, so she decides to eliminate such effects by making sure to record each store's sales on the same sample of days. After choosing a random sample of 8 days, she records the sales (in dollars) for each store on these days, as shown in the table below. Day 1 2 3 4 5 6 7 8 Store 1 654 303 989 591 872 934 828 947 Store 2 498 114 674 605 912 1008 686 848 Difference 156 189 315 - 14 - 40 - 74 142 99 (Store 1 - Store 2) Send data to calculator Based on these data, can the owner conclude, at the 0.05 level of significance, that the mean daily sales of the two stores differ? Answer this question by performing a hypothesis test regarding u, (which is u with a letter "d" subscript), the population mean daily sales difference between the two stores. Assume that this population of differences…
Answer these questions A, B, C and D question
You are conducting a multinomial hypothesis test (α= 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. Category ObservedFrequency ExpectedFrequency A 17 B 22 C 11 D 13 E 17 Report all answers accurate to three decimal places. But retain unrounded numbers for future calculations.What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places, and remember to use the unrounded Pearson residuals in your calculations.)χ2= What are the degrees of freedom for this test?d.f.= What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value = Is the p-value less than αα? yes no This test statistic leads to a decision to... reject the null accept the null fail to reject the null accept the alternative As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to…
You are conducting a multinomial hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Category ObservedFrequency ExpectedFrequency SquaredPearsonResidual A 5 B 12 C 24 D 5 E 8 What is the chi-square test-statistic for this data?χ2=χ2= Report all answers accurate to three decimal places.
A personal computer manufacturer is interested in comparing assembly times for two keyboard assembly processes. Assembly times can vary considerably from worker to worker, and the company decides to eliminate this effect by selecting a random sample of 12 workers and timing each worker on each assembly process. Half of the workers are chosen at random to use Process 1 first, and the rest use Process 2 first. For each worker and each process, the assembly time (in minutes) is recorded, as shown in the table below. Worker 1 2 3 4 5 6. 7 8. 9. 10 11 12 Process 1 85 63 46 85 83 70 64 51 48 69 67 52 Process 2 88 46 43 61 80 51 59 53 23 46 59 52 Difference - 3 17 3 24 19 5 -2 25 23 8 (Process 1 - Process 2) Send data to calculator Based on these data, can the company conclude, at the 0.05 level of significance, that the mean assembly times for the two processes differ? Answer this question by performing a hypothesis test regarding u, (which is with a letter "d" subscript), the population…
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Consider the data. xi 3 12 6 20 14 yi 60 40 65 5 15 Use the t test to test the following hypotheses (? = 0.05): H0: ?1 = 0 Ha: ?1 ≠ 0 Find the value of the test statistic. (Round your answer to three decimal places.) Use the F test to test the hypotheses in part (d) at a 0.05 level of significance. Present the results in the analysis of variance table format.
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P-value is -.055 You are asked to make a decision about whether to reject the null hypothesis that the population value of gamma equals 0 (i.e., that there is no ordinary association in the population between frequency of prayer and health). If alpha = 0.05 and you use a non-directional alternative hypothesis, which of the following is true?) A. Since the p-value is larger than alpha, you do not reject the null hypothesis B. Since the p-value is less than alpha, you do not reject the null hypothesis C. Since the p-value is larger than alpha, you reject the null hypothesis. D. Since the p-value is less than alpha, you reject the null hypothesis
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State whether the standardized test statistic t indicates that you should reject the null hypothesis. Explain. (a) t= 1.584 (b) t=0 (c) t= - 1.526 (d) t= - 1.604 to = - 1.558 (a) For t= 1.584, should you reject or fail to reject the null hypothesis? O A. Fail to reject Ho, because t> - 1.558. O B. Reject Ho, because t> - 1.558. O C. Reject Ho, because t - 1.558. O B. Fail to reject Ho, because t - 1.558. (c) For t= - 1.526, should you reject or fail to reject the null hypothesis? O A. Reject Ho, because t - 1.558. O C. Fail to reject Ho, because t> - 1.558. O D. Fail to reject Hn, because t< - 1.558.