EBK AN INTRODUCTION TO MATHEMATICAL STA
6th Edition
ISBN: 9780134114248
Author: Marx
Publisher: YUZU
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Chapter 6.4, Problem 12Q
(a)
To determine
To calculate: The value of
(b)
To determine
To calculate: The value of
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A manufacturing company employs two inspecting devices to sample a fraction of their output for quality control purposes. The first inspection monitor is able to accurately detect 99.3% of the defective items it receives, whereas the second is able to do so in 99.7% of the cases. Assume that four defective items are produced and sent out for inspection. Let X and Y denote the number of items that will be identified as defective by inspecting devices 1 and 2, respectively. Determine the following.
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1) fxy(x,y) 2) fx (x)3) fY|2 4) E(x)Show all work
Chapter 6 Solutions
EBK AN INTRODUCTION TO MATHEMATICAL STA
Ch. 6.2 - State the decision rule that would be used to test...Ch. 6.2 - An herbalist is experimenting with juices...Ch. 6.2 - (a) Suppose H0:=0 is rejected in favor of H1:0 at...Ch. 6.2 - Company records show that drivers get an average...Ch. 6.2 - If H0:=0 is rejected in favor of H1:0, will it...Ch. 6.2 - A random sample of size 16 is drawn from a normal...Ch. 6.2 - Recall the breath analyzers described in Example...Ch. 6.2 - Calculate the P-values for the hypothesis tests...Ch. 6.2 - Suppose H0:=120 is tested against H1:120. If =10...Ch. 6.2 - As a class research project, Rosaura wants to see...
Ch. 6.2 - As input for a new inflation model, economists...Ch. 6.3 - Commercial fishermen working certain parts of the...Ch. 6.3 - Efforts to find a genetic explanation for why...Ch. 6.3 - Defeated in his most recent attempt to win a...Ch. 6.3 - Suppose H0:p=0.45 is to be tested against H1:p0.45...Ch. 6.3 - Recall the median test described in Example 5.3.2....Ch. 6.3 - Among the early attempts to revisit the death...Ch. 6.3 - What levels are possible with a decision rule of...Ch. 6.3 - Suppose H0:p=0.75 is to be tested against H1:p0.75...Ch. 6.4 - Recall the Math for the Twenty-First Century...Ch. 6.4 - Carry out the details to verify the decision rule...Ch. 6.4 - For the decision rule found in Question 6.2.2 to...Ch. 6.4 - Construct a power curve for the =0.05 test of...Ch. 6.4 - If H0:=240 is tested against H1:240 at the =0.01...Ch. 6.4 - Suppose n=36 observations are taken from a normal...Ch. 6.4 - If H0:=200 is to be tested against H1:200 at the...Ch. 6.4 - Will n=45 be a sufficiently large sample to test...Ch. 6.4 - If H0:=30 is tested against H1:30 using n=16...Ch. 6.4 - Suppose a sample of size 1 is taken from the pdf...Ch. 6.4 - Polygraphs used in criminal investigations...Ch. 6.4 - Prob. 12QCh. 6.4 - Prob. 13QCh. 6.4 - A sample of size 1 is taken from the pdf...Ch. 6.4 - Prob. 15QCh. 6.4 - Prob. 16QCh. 6.4 - Prob. 17QCh. 6.4 - Prob. 18QCh. 6.4 - Prob. 19QCh. 6.4 - Suppose that one observation from the exponential...Ch. 6.4 - Prob. 21QCh. 6.4 - Prob. 22QCh. 6.4 - Prob. 23QCh. 6.4 - Given the pdf fY(y;)=2y2,0y. Take a sample of size...Ch. 6.5 - Let k1,k2,...,kn be a random sample from the...Ch. 6.5 - Let y1,y2,...,y10 be a random sample from an...Ch. 6.5 - Let y1,y2,...,yn be a random sample from a normal...Ch. 6.5 - Let k denote the number of successes observed in a...Ch. 6.5 - Suppose a sufficient statistic exists for the...
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- A manufacturing company employs two inspecting devices to sample a fraction of their output for quality control purposes. The first inspection monitor is able to accurately detect 99.3% of the defective items it receives, whereas the second is able to do so in 99.7% of the cases. Assume that four defective items are produced and sent out for inspection. Let X and Y denote the number of items that will be identified as defective by inspecting devices 1 and 2, respectively. Determine the following. 1) fxy(x,y) 2) fx (x)3) E(x)4) E(Y|X=2)5) V(Y|X=2)6) Are X and Y independent? Why?arrow_forwardIn testing the hypotheses H0: p = 0.5 vs Ha: p <= 0.5, suppose you get a p-value 0.022. Thenyou realize that the one sided alternative is too restrictive and re-do the test with a two sidedalternative Ha : p =/= 0.5.Based on the new p-value, what will be your conclusion ? A. Reject H0 at alpha = 0.01 but not at alpha = 0.05; 0.1.B. Reject H0 at alpha = 0.05 and alpha = 0.1 but not at alpha = 0.01.C. Fail to reject H0 at all the above alpha values.D. Reject H0 at alpha = 0.05 and alpha = 0.01 but not at alpha = 0.1.E. Reject H0 at all the above alpha values.arrow_forward
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