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A random sample of n observations is selected from a normal population to test the null hypothesis that σ2=25. Specify the rejection region for each of the following combinations of Ha,αand n.

a.Ha:σ225;α=0.5;n=16

b.Ha:σ2>25;α=.10;n=15

c.Ha:σ2>25;α=.01;n=23

d. Ha:σ2<25;α=.01;n=13

e. Ha:σ225;α=.10;n=7

f. Ha:σ2<25;α=.05;n=25

Short Answer

Expert verified

a. χ2<6.26214or χ2>27.4884

b. χ2>40.2894

c. χ2>21.0642

d. χ2<3.57056

e. χ2<1.63539or χ2>12.5916

f.χ2<13.8484

Step by step solution

01

Defining the Rejection Region

Rejection Region holds significance in the sense that whenever a numerical value of the test statistic falls in the rejection region, the null hypothesis is rejected.

02

Solving for part a.

Chi-square distribution depends on (n-1)degrees of freedom. With α=0.05and (n-1)=15, the χ2value for rejection is found in Table IV of Appendix D.

Since this is a two-tailed test, therefore we have χ2<6.26214or χ2>27.4884which are the values for the two tailed rejection region.

03

Solving for part b.

With α=0.01,n=23, we have (n-1)=22.

Therefore, the rejection value will be χ2>40.2894.

04

Solving for part c.

With α=0.10,n=15, we have (n-1)=14.

Therefore, the rejection value will be χ2>21.0642.

05

Solving for part d.

With α=0.01,n=13, we have (n-1)=12

Therefore, the rejection value will be χ2<3.57056.

06

Solving for part e.

With α=0.10,n=7, we have (n-1)=6.

Since this is a two-tailed test, therefore the values for rejection will be χ2<1.63539or χ2>12.5916.

07

 Step 7: Solving for part f.

With α=0.05,n=25, we have (n-1)=24.

Therefore, the rejection value will beχ2<13.8484.

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Most popular questions from this chapter

Non-destructive evaluation. Non-destructive evaluation(NDE) describes methods that quantitatively characterize materials, tissues, and structures by non-invasive means, such as X-ray computed tomography, ultrasonic, and acoustic emission. Recently, NDE was used to detect defects in steel castings (JOM,May 2005). Assume that the probability that NDE detects a “hit” (i.e., predicts a defect in a steel casting) when, in fact, a defect exists is .97. (This is often called the probability of detection.) Also assume that the probability that NDE detects a hit when, in fact, no defect exists is .005. (This is called the probability of a false call.) Past experience has shown a defect occurs once in every 100 steel castings. If NDE detects a hit for a particular steel casting, what is the probability that an actual defect exists?

Assume that x is a binomial random variable with n = 1000 andp = 0.50. Use a normal approximation to find each of the following probabilities:

a. P(x>500)

b.P(490x<500)

c.P(x>550)

A random sample of n = 6 observations from a normal distribution resulted in the data shown in the table. Compute a 95% confidence interval for σ2

Question: The purpose of this exercise is to compare the variability of with the variability of .

a. Suppose the first sample is selected from a population with mean and variance . Within what range should the sample mean vary about of the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations of on each side of .

b. Suppose the second sample is selected independently of the first from a second population with mean and variance . Within what range should the sample mean vary about the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations on each side .

c. Now consider the difference between the two sample means . What are the mean and standard deviation of the sampling distribution ?

d. Within what range should the difference in sample means vary about the time in repeated independent samples of measurements each from the two populations?

e. What, in general, can be said about the variability of the difference between independent sample means relative to the variability of the individual sample means?

Question: Refer to the Bulletin of Marine Science (April 2010) study of lobster trap placement, Exercise 6.29 (p. 348). Recall that the variable of interest was the average distance separating traps—called trap-spacing—deployed by teams of fishermen. The trap-spacing measurements (in meters) for a sample of seven teams from the Bahia Tortugas (BT) fishing cooperative are repeated in the table. In addition, trap-spacing measurements for eight teams from the Punta Abreojos (PA) fishing cooperative are listed. For this problem, we are interested in comparing the mean trap-spacing measurements of the two fishing cooperatives.

BT Cooperative

93

99

105

94

82

70

86

PA Cooperative

118

94

106

72

90

66

98


Source: Based on G. G. Chester, “Explaining Catch Variation Among Baja California Lobster Fishers Through Spatial Analysis of Trap-Placement Decisions,” Bulletin of Marine Science, Vol. 86, No. 2, April 2010 (Table 1).

a. Identify the target parameter for this study.b. Compute a point estimate of the target parameter.c. What is the problem with using the normal (z) statistic to find a confidence interval for the target parameter?d. Find aconfidence interval for the target parameter.e. Use the interval, part d, to make a statement about the difference in mean trap-spacing measurements of the two fishing cooperatives.f. What conditions must be satisfied for the inference, part e, to be valid?

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