The following MINITAB output presents the results of a hypothesis test for the differenceP1 - P2 between two population proportions. Some of the numbers are missing. Fill in thenumbers for (a) through (d).Test and CI for Two ProportionsSample p(a)Sampleх1011532(b)900.544444Difference = p (1) - (2)Estimate for difference: 0.11568695% CI for difference: (-0.0116695, 0.243042)Test for difference = 0 (vs not = 0): Z = (c) P-Value = (d)%3D%3D

The values missed in the table will be obtained as given below: (a)

Answered: The following MINITAB output presents…

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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Three runs of an assay generate 72.49%, 72.99% and 72.40%. The average result needs to be accurate to 1%. What is the proper expression for average?
Given a 95% C.I. = (34.14 mm, 36.02 mm) for a sample of Wooly Bear caterpillar lengths (preceding problem), address the following. Based upon the above information, I am _____ % confident that the 95% C.I. contains the sample mean.
From the table on the image: Construct an appropriate hypothesis test to determine whether or not there is an association between experiencing phenomena such as seeing a ghost and the age group you are in. Write the appropriate null/alternative hypothesis and carry out the test. Use a significance level of α = 0.01.
Given SST = 2415, SSE = 1725, n = 7, and k = 2, calculate the value of the overall F test statistic. O Feal = 3.50 O Fcal = 1.25 O Fcal = 0.40 O Fcal = 0.80
LSU's Ag. Center wants to determine whether the mean yield for sugarcane is different this year compared to the historical avg of 8,700 lbs of sugar per acre. Thus, the null and alternative hypotheses are Ho: mu = 8,700 Ha: mu is not = to 8,700 You set out to conduct a hypothesis test at the 95% level. Mean yields are distributed Normally. You take a sample of 100 different acres. From this sample, you calculate a mean yield of 8,853 and a std deviation of 1,150. What are the bounds for the acceptance region for this hypothesis test? Group of answer choices -1.64 to 1.64 -2.58 to 2.58 -1.33 to 1.33 -1.96 and 1.96
Data for gas mileage (in mpg) for different vehicles was entered into a software package and part of the ANOVA table is shown below: Source DF SS MS Vehicle 5 420 210.00 Error 66 321 4.86 Total 71 741 If a LSRL was fit to this data, what would the value of the coefficient of determination be?
Epistasis is a phenomenon in which one gene controls the expression of another. In one case, a dihybrid cross should yield a 9:3:4 ratio. You collect the following data on mice: black coat: 138 brown coat: 46 white coat: 60 Is there any reason to doubt the ratio of 9 black : 3 brown : 4 white?
instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 63.9 68.4 | 69 | 96 64.3 | 63.9 65 69 62.7 68.4 Test grades are believed to be normally distributed. Use a significance level of 5%. A. State the alternative hypothesis: HA: Ou + 65 Ομ 65 B. State the mean of the sample: C. State the standard error of the sample means:
(3-B is needed to answer it I am only asking for question 4.) 3) You go out and collect the following estimates of earthworms / acre:54,276 57,378 51,108 66,190 66,232 59,018 57,159These data yield the following: ̄y = 59,765.86, and s = 5,689.606a) Construct a 90% CI for these data.b) Construct a 99% CI for these data.c) Darwin once estimated that an acre of soil had about 50,000 worms in it. Is his estimateconsistent with the data above? (Historical note: His estimate was considered way too high inhis day). 4) Consider the results of 3(b). Notice that all the data fit within the 99% confidence interval. Is thisusually the case (in other words, will a 99% CI contain most of the observations)?Caution: a lot of people get this wrong! Here's a hint: suppose you had measured the worms in6000 acres (instead of just 7). What happens to the confidence interval? If you're not sure,substitute 6,000 for 7 in your calculation for (b) to see what happens.
If the test statistic for a right sided test is z = 1.71, then the p-value isSelect one: A.0.9573 B.0.9564 C.0.0436 D.0.0427
a survey is conducted to compare the proportions of men and women whoaccess their bank statements online. Denoting the population propotions by Pm and Pw, a two-proportion z-test is carried out to test Ho: pm = pw against Ha: pm > pw. The values of the test statistic is found to be z = 0.784, and the p-value for the test is found to be 0.216. Which of the following is the correct interpretation of the p-value? a) given that ha is true, the probability of getting a value of z at least as large as 0.784 is 0.216. b) given that ho is true, the probability of getting a value of z at least as large as 0.784 is 0.216. c) given the results of the survey, the probability that Pm > Pw is 0.216. d) given the results of the survey, the probability that Pm = Pw is 0.216. e) none of the above
Only answer the questions typed: A. indicate whether it's one or two sidedB. Indicate whether it's a test of means or proportionsC. Write out the expression for determining the P-value

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