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Let t0 be a particular value of t. Use Table III in Appendix D to find t0 values such that the following statements are true.

a.=P(-t0<t<t0).95wheredf=10b.P(t-t0ortt0)wheredf=10c.P(tt0)=.05wheredf=10d.P(t-t0ortt0)=.10wheredf=20e.P(t-t0ortt0)=.01wheredf=5

Short Answer

Expert verified

Answer

  1. Not true.
  2. Not applicable
  3. True
  4. True
  5. True

Step by step solution

01

Determining the value of t0 when the probability is 0.95 and degrees of freedom is 10

a.

The two-tail test is applicable in this case. With the help of MS Excel, the exact value of t0 can be found to be 0.064298 when the degrees of freedom is 10. This shows that statement is not true. The value of t0 is very less than the values of t (taking reference from Appendix D).

02

Determining the value of t0 when the probability is unknown and degrees of freedom is 10

In this subpart, the solution cannot be found as the value of the probability is not there.The value of the probability is required in this subpart to denote the value of the t0.

03

Determining the value of t0 when the probability is 0.05 and degrees of freedom is 10

c.

In this subpart, the one-tail test is applicable. With the help of MS Excel, the exact value of t0 can be found to be 1.81246 when the degrees of freedom is 10. This shows that statement is true. The value of t0 is greater than the value of t at t.050, which is 1.725 (taking reference from Appendix D).

04

 Step 4: Determining the value of t0 when the probability is 0.10 and degrees of freedom is 20

e.

In this subpart, the two-tail test is applicable, and so, with the help of MS Excel, the exact value of t0 can be found to be 1.72472 when the degrees of freedom is 20. This shows that statement is true as the value of t0 is greater than the value of t at t.100, which is 1.325 (taking reference from Appendix D).

05

Determining the value of t0 when the probability is 0.01 and degrees of freedom is 5

f.

In this subpart, the two-tail test is applicable, and so, with the help of MS Excel, the exact value of t0 can be found to be 4.0321 when the degrees of freedom is 5. This shows that statement is true as the value of t0 is greater than the value of t at t.010, which is 3.365 (taking reference from Appendix D).

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Most popular questions from this chapter

Question: Forecasting daily admission of a water park. To determine whether extra personnel are needed for the day, the owners of a water adventure park would like to find a model that would allow them to predict the day’s attendance each morning before opening based on the day of the week and weather conditions. The model is of the form

where,

y = Daily admission

x1 = 1 if weekend

0 otherwise

X2 = 1 if sunny

0 if overcast

X3 = predicted daily high temperature (°F)

These data were recorded for a random sample of 30 days, and a regression model was fitted to the data.

The least squares analysis produced the following results:

with

  1. Interpret the estimated model coefficients.
  2. Is there sufficient evidence to conclude that this model is useful for predicting daily attendance? Use α = .05.
  3. Is there sufficient evidence to conclude that the mean attendance increases on weekends? Use α = .10.
  4. Use the model to predict the attendance on a sunny weekday with a predicted high temperature of 95°F.
  5. Suppose the 90% prediction interval for part d is (645, 1,245). Interpret this interval.

Identify the rejection region for each of the following cases. Assume

v1=7andv2=9

a. Ha1222,α=.05

b. Ha1222,α=.01

c. Ha12σ22,α=.1withs12>s22

d. Ha1222,α=.025

The “winner’s curse” in transaction bidding. In transaction bidding, the “winner’s curse” is the miracle of the winning (or loftiest) shot price being above the anticipated value of the item being auctioned. The Review of Economics and Statistics (Aug. 2001) published a study on whether shot experience impacts the liability of the winner’s curse being. Two groups of a stab in a sealed-shot transaction were compared (1)super-experienced stab and (2) less educated stab. In the super-experienced group, 29 of 189 winning flings were above the item’s anticipated value; 32 of 149 winning flings were above the item’s anticipated value in the less-educated group.

  1. Find an estimate of p1, the true proportion of super educated stab who fell prey to the winner’s curse
  2. Find an estimate of p2, the true proportion of less-educated stab who fell prey to the winner’s curse.
  3. Construct a 90 confidence interval for p1-p2.
  4. d. Give a practical interpretation of the confidence interval, part c. Make a statement about whether shot experience impacts the liability of the winner’s curse being.

Question:Quality control. Refer to Exercise 5.68. The mean diameter of the bearings produced by the machine is supposed to be .5 inch. The company decides to use the sample mean from Exercise 5.68 to decide whether the process is in control (i.e., whether it is producing bearings with a mean diameter of .5 inch). The machine will be considered out of control if the mean of the sample of n = 25 diameters is less than .4994 inch or larger than .5006 inch. If the true mean diameter of the bearings produced by the machine is .501 inch, what is the approximate probability that the test will imply that the process is out of control?

Two populations are described in each of the following cases. In which cases would it be appropriate to apply the small-sample t-test to investigate the difference between the population means?

a.Population 1: Normal distribution with variance σ12. Population 2: Skewed to the right with varianceσ22=σ12.

b. Population 1: Normal distribution with variance σ12. Population 2: Normal distribution with variance σ22σ12.

c. Population 1: Skewed to the left with variance σ12. Population 2: Skewed to the left with varianceσ22=σ12.

d. Population 1: Normal distribution with varianceσ12 . Population 2: Normal distribution with varianceσ22=σ12 .

e. Population 1: Uniform distribution with varianceσ12 . Population 2: Uniform distribution with variance σ22=σ12.

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