A uniform distribution from 3.6 to 8.6 has a mean E[X] of, O 6.10 O 3.05 O 12.20 O 2.50
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![A uniform distribution from 3.6 to 8.6 has a mean E[X] of ,
O 6.10
O 3.05
O 12.20
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- An employment information service claims the mean annual salary for senior level product engineers is $97,000. The annual salaries (in dollars) for a random sample of 16 senior level product engineers are shown in the table to the right. At α=0.10, test the claim that the mean salary is $97,000. Complete parts (a) through (e) below. Assume the population is normally distributed. Annual Salaries 100 ,637 96 ,395 93 ,509 112 ,609 82 ,472 74 ,162 76 ,989 80 ,977 102 ,467 76 ,251 103 ,944 104 ,042 91 ,135 82 ,152 85 ,096 110 ,269 Part 1 (a) Identify the claim and state H0 and Ha. (Type integers or decimals.) Part 2 (b) Use technology to find the critical value(s) and identify the rejection region(s). The critical value(s) is/are t0= (c) Find the standardized test statistic, t. The standardized test…APEAS of a Standard Normal Distribution (b) Tuble of Areas to the Left of z When z Is Pusitive .00 .01 .02 .03 .04 .05 .06 .07 ,08 .09 (a) Table of Areas to the Lefi of z When z Is Negative 0,0 5000 .5040 5080 .5120 5160 5199 .5239 5279 .5319 .5359 .00 .01 03 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 02 .04 05 06 .07 08 09 0.1 5398 .5438 5478 .5517 5557 5596 .5636 5675 .5714 .5753 .0002 0.2 .5793 .5832.5871 5910 .5948 5987 .6026 ,6064 6103 .6141 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 s the areu to the .0003 0.3 .6179 ,6217 ,6255 .6293 .6331 .6368 .6406 6443 .6480 .6517 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 0,4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 -3.1 ,0010 ,0009 0009 .0009 .0008 .0008 .0008 ,0008 .0007 „6915 6950 6985 .7019 .7054 .7088 .7123 .7157 7190 .7224 0.5 .0007 -3.0 .0013 ,0013. .0013 .0012 .0012 0011 .0011 .0011 .0010 .0010 0.6 .7257 7291 .7324 .7357 .7389 .7422 .7454 '.7486 .7517 .7549 -2.9…Suppose that X-bin(65,0.2). Using the normal approximation to the binomlal distribution, P(12.< X < 14.) is closest to: O a. 0.30 O b.0.26 Oc.0.36 O d. 0.24 O e. 0.18 ity Jump to... 21
- 1.)Find the z-score corresponding to the given area of the normal curve with the following given values: a.) Area to the right of -z is .6736 b.) Area to the left of z is .75183. For the binomial distribution with the given values for n and p, state whether or not it is suitable to use the normal distribution as an approximation. n= 65 and p = 0.7 a. Normal approximation is not suitable. b. Normal approximation is suitable.4. If T follows a 17 distribution, find to such that (a) P(T| to) = .05.
- Are couples that live together before they get married more likely to end up divorced within five years of marriage compared to couples that live apart before they get married? 217 of the 671 couples from the study who lived together before they got married were divorced within five years of marriage. 132 of the 421 couples from the study who lived apart before they got married were divorced within five years of marriage. What can be concluded at the = 0.10 level of significance? For this study, we should use The null and alternative hypotheses would be: 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 Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The results are statistically insignificant at αα = 0.10, so there is statistically significant evidence to conclude that the divorce rate…To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material. 3 1 19 21 25 20 23 24 21 23 a. Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use a 0.05. Compute the values below (to 2 decimals, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). Manufacturer What conclusion can you draw after carrying out this test?. Select your answer- 2 29 27 32 28 The p-value is -Select your answer What is your conclusion? -Select your answer- b. At the a 0.05 level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers 1 and 3. Calculate Fisher's LSD Value (to 2 decimals).3) During a study, the weights of test subjects was recorded. The results are displayed in the graph below. Use the graph to determine the value of the following descriptive statistics Histogram of Wt (n=40) 20 15 10 10 100 150 X Values D How many weights were recorded? D) What is the average weight? O Approximate the standard deviation. d) What is the one-standard deviation interval around the mean? How many weights were within this interval? I What percentage of weights were within this interval? What is the two standard deviation interval around the mean? hi How many weights werE within this interval? What percentage of we were within thisterva What is the three standard deviation ntervalarund the mean K) How many weights were within this interv What percentage of weights were withiin thikinterval? 梦 $ hp
- A particular paper included the accompanying data on the tar level of cigarettes smoked for a sample of male smokers who subsequently died of lung cancer. Assume it is reasonable to regard the sample as representative of male smokers who die of lung cancer. Is there convincing evidence that the proportion of male smoker lung cancer deaths is not the same for the four given tar level categories at the ? = .05 level? (Use 2 decimal places.) Tar Level Frequency 0-7 129 8-14 324 15-21 543 22 153 ?2 = P-value interval p < 0.0010.001 ≤ p < 0.01 0.01 ≤ p < 0.05 0.05 ≤ p < 0.10 p ≥ 0.10Suppose the random variable Z follows a standard normal distribution,, Then thevalue of PIZ<-339) is O A 0.9997 O B. 0,0003 O C. 1.000 O D. 0.0000Q9. Let X,, X2, .., X, be a random sample a gamma distribution with a= 3 and B = 0>0, i.e., ..... with population p.d.f. f(x; 0) =x?e. Show that I. the MLE of e is ên = X/3. II. ê = X/3 is unbiased estimator of 0. III. ê = X/3 is MVUE of 0. %3D
