Question 1041095: A bond analyst is analyzing the interest rates for equivalent municipal bonds issued by two different states. At α = 0.05, is there enough evidence to conclude that there is a difference in the interest rates paid by the two states?
State A
State B
Sample size
70
70
Mean interest rate (%)
3.6
4.0
Population variance
0.04
0.05
Yes, because the test value –11.16 is outside the critical region
Yes, because the test value –3.11 is outside the critical region
Yes, because the test value 124.44 is outside the critical region
No, because the test value –0.01 is inside the critical region
A marketing firm asked a random set of married and single men how much they were willing to spend on a vacation. Is there sufficient evidence at α = 0.05 to conclude that is there a difference in the two amounts? (Hint: This is z test of two means, follow the procedure in Section 9 - 1, variance is sigma squared )
Married men
Single men
Sample size
40
50
Sample mean
$920
$845
Population variance
5700
9100
No, because the test value 0.23 is inside the critical region .
Yes, because the test value 4.16 is outside the critical region .
Yes, because the test value 1.60 is inside the critical region .
No, because the test value 1.60 is outside the critical region .
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QUESTION 4
If the probability of a type II error in a hypothesis test is 0.25, and 0.05, then the power of this test is
0.05
0.25
0.75
0.95
According to Beautiful Bride magazine, the average age of a groom is now 26.2 years. A sample of 16 prospective grooms in Chicago revealed that their average age was 26.6 years with a standard deviation of 5.3 years. What is the test value for a t test of the claim? (Hint: Since the population standard deviation is unknown use t test value, section 8 - 3)
1.81
0.30
0.59
2.13
A recent survey indicated that the average amount spent for breakfast by business managers was $7.58 with a standard deviation of $0.42. It was felt that breakfasts on the West Coast were higher than $7.58. A sample of 81 business managers on the West Coast had an average breakfast cost of $7.65. (Hint: Find the z test value and P - value is a tail probability for this one tailed test)
Find the P-value for the test.
0.4332
0.2734
0.1325
0.0668
For a random sample of 23 European countries, the variance on life expectancy was 7.3 years. What is the 95% confidence interval for the variance of life expectancy in all of Europe?
An economics professor randomly selected 100 millionaires in the United States. The average age of these millionaires was 54.8 years. If the standard deviation of the entire population of millionaires is 7.9 years, find the 95% confidence interval for the mean age of all United States millionaires. (Hint: Since the population standard deviation is known use z confidence interval)
A lumber mill is tested for consistency by measuring the variance of board thickness. The target accuracy is a variance of 0.0035 square inches or less. If 30 measurements are made and their variance is 0.006 square inches, is there enough evidence to reject the claim that the standard deviation is within the limit at = .01? (Hint: Chi squre test of a variance)
Yes, since the χ2 test value 49.71 is greater than the critical value 49.588.
No, since the χ2 test value 9.23 is less than the critical value 49.588.
Yes, since the χ2 test value 49.71 is less than the critical value 50.892.
No, since the χ2 test value 9.23 is less than the critical value 50.892.
68% of students at a university live on campus. A random sample found that 24 of 40 male students and 43 of 55 of female students live on campus. At the 0.05 level of significance, is there sufficient evidence to support the claim that a difference exists between the proportions of male and female students who live on campus? (Hint: This is z test of two proportions, follow the procedure in Section 9 - 4)
Yes, because the test value –20.26 is outside the noncritical region –1.96 < z < 1.96.
No, because the test value –0.96 is inside the noncritical region –1.96 < z < 1.96.
Yes, because the test value –4.21 is outside the noncritical region –1.96 < z < 1.96.
No, because the test value –1.92 is inside the noncritical region –1.96 < z < 1.96.
3 squirrels were found to have an average weight of 9.3 ounces with a sample standard deviation is 1.1. Find the 95% confidence interval of the true mean weight. (Hint: Since the population standard deviation is unknown use t-distribution with 2 degrees of freedom)
A scientist claims that only 65% of geese in his area fly south for the winter. He tags 55 random geese in the summer and finds that 17 of them do not fly south in the winter. If = .05, is the scientist's belief warranted? (Hint: Z test of proportion)
Yes, because the test value 0.64 is in the noncritical region.
No, because the test value 0.70 is in the critical region.
Yes, because the test value –0.70 is in the noncritical region.
No, because the test value –0.64 is in the noncritical region.
A recent study of 750 internet users in Europe found that 35% of internet users were women. What is the 95% confidence interval of the true proportion of women in Europe who use the internet? (Hint: Confidence interval proportion, use z confidence interval, Section 7 - 3)
What is the value for for a 95% confidence interval when n = 18? (Hint: for 95 % confidence each tail should be 2.5 %)
7.564
8.672
9.390
8.231
A recent survey reported that in a sample of 300 students who attend two-year colleges, 105 work at least 20 hours per week. Additionally, in a sample of 225 students attending private four-year universities, only 20 students work at least 20 hours per week. What is the test value for a test of the difference between these two population proportions? (Hint: Section 9.4)
6.95
7.61
2.38
4.18
Using the z table, determine the critical values for a two-tailed test when α = 0.03.
± 1.88
± 2.17
± 0.18
± 0.06
Answer by ikleyn(52781)   (Show Source):
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