Download 1 ISC 207 Practice questions for Quiz-3 (Solutions)

ISC 207
Practice questions for Quiz-3 (Solutions)
1. According to the Central Limit Theorem, the sampling distribution of the mean has a normal
distribution for large samples sizes (n >= 30) even if the underlying distribution is not normal. T / F
True
2. Suppose a random variable is distributed normally with a mean of 42 and a std. dev. of 12. The
sampling distribution of the mean of samples of size 36 will have a mean and std. dev. of:
a. 42,12
b. 7,12
c. 7,2
d. 42,36
e. 42,2
mean is the same, std. dev. is 12/sqrt(36) = 2
3. As the sample size increases, the standard deviation of sample mean decreases. T / F True, because
the standard deviation of sample mean is σ/sqrt(n). So as n increases, the std. dev. decreases.
4. As the confidence level increases, the size of the confidence interval decreases (For the same samp le
size). T / F
False, because CI = X-bar ± zα/2 . σ/sqrt(n)
Now, as the confidence level increases, α decreases and zα/2 increases, therefore CI increases.
5. For the same confidence level, as the sample size increases, the size of the confidence interval
decreases. T / F True, because CI = X-bar ± zα/2. σ/sqrt(n). Since n is in the denominator, as n
increases, CI decreases.
6. Suppose I want to estimate the mean household income of a neighborhood within a city. I get a
sample of 50 households and ask them of their household income. The mean household income for
this sample of 50 households in US dollars was 66,000. The standard deviation was 12,000.
a. What is a 90% confidence interval for the mean household income for the neighborhood?
66,000 ± z0.05 *12,000/sqrt(50) = 66,000 ± 1.645*12,000/7.071 = 66000 ± 2791
=[63209, 68791]
b. What is a 95% confidence interval?
66,000 ± z0.025 *12,000/sqrt(50) = 66,000 ± 1.96*12,000/7.071 = 66000 ± 3326
=[62674, 69326]
c. What is a 98% confidence interval?
66,000 ± z0.01 *12,000/sqrt(50) = 66,000 ± 2.33*12,000/7.071 = 66000 ± 3954
=[62046, 69954]
7. Suppose in the above problem, the sample size was only 18 households. All other information is the
same.
a. What is a 90% confidence interval for the mean household income for the neighborhood?
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66,000 ± T0.05 *12,000/sqrt(18) = 66,000 ± 1.740*12,000/4.2426 = 66000 ± 4921=
=[61079, 70921]
Note: 1.74 is T of 0.05 for 17 dof.
b. What is a 95% confidence interval?
66,000 ± T0.025*12,000/sqrt(18) = 66,000 ± 2.110*12,000/4.2426 = 66000 ± 5968=
[60032, 71968]
Note: 2.110 = t of 0.025 at 17 dof.
c. What is a 98% confidence interval?
66,000 ± T0.01*12,000/sqrt(18) = 66,000 ± 2.567*12,000/4.2426 = 66000 ± 7260=
[58740, 73260]
Note: 2.567 is t of 0.01 at 17 dof
8. How do you interpret a 99% CI of [20,25]?
We are 99% confident that the true population mean lies between 20 and 25.
9. Suppose H0: µ = 100 and Ha: µ < 100, below what vale of X-bar will we reject H0 if we want a desired
significance level of 0.05, assuming that we have a sample of size 81 and a sample standard deviation
of 18?
This is a one-tailed (left tailed) hypothesis.
So the question is what is X such that P(X-bar <= X) = 0.05. We assume that the distribution of X-bar
is normal with mean of 100 and standard deviation of 18/Sqrt(81) or 2.
Since P(z<-1.645)=0.05, at 1.645 or more standard deviation below the mean, we will reject H0.
1.645 std. dev. = 1.645*2 = 3.29. So at 100 – 3.29 or below we will reject H0 or at 96.71 or below we
will reject H0.
10. Do the above problem, with the only difference that the sample size is 16 instead of 81.
In this problem, std. dev. of the mean is 18/4 = 4.5
Using the T-table we find that for 15 dof, P(T<1.753) = 0.05.
So at 1.753 std. dev. below the mean, we will reject H0.
1.753*4.5 = 7.89
So, we will reject H0 at 92.11 or below.
11. Suppose H0: µ = 100 and Ha: µ ≠ 100, at what vale of X-bar will we reject H0 if we want a desired
significance level of 0.05, assuming that we have a sample of size 81 and a sample standard deviation
of 18?
This is a two-tailed hypothesis. H0 will be rejected if X-bar is either too high or too low.
At +1.96 or more std. dev. or -1.96 or less std. dev. we will reject the Null.
1.96 std dev. = 1.96 * 18/9 = 1.96*2 = 3.92. So at 103.92 or above or 96.08 or below we will reject
H0.
12. Do the above problem, with the only difference that the sample size is 16 instead of 81.
Using the T-table, we find that for 15 dof, P(T>0.025)=2.132
At +2.132 or more std. dev or -2.132 or less std. dev. we will reject the Null.
2.132 std dev = 2.132*4.5 = 9.59. So at 109.59 or more or 90.41 or below we will reject H0
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13. Suppose H0: µ = 100 and Ha: µ > 100, over what vale of X-bar will we reject H0 if we want a desired
significance level of 0.05, assuming that we have a sample of size 16 and a sample standard deviation
of 18?
This is a one-tailed (right tailed) hypothesis.
So the question is what is X such that P(X-bar >= X) = 0.05. We will assume that the distribution of Xbar is normal with mean of 100 and standard deviation of 18/Sqrt(81) or 2.
Since P(z>1.645)=0.05, at 1.645 or more standard deviation above the mean, we will reject H0.
1.645 std. dev. = 1.645*2 = 3.29. So at 100 + 3.29 or above we will reject H0 or at 103.29 or above
we will reject H0.
14. Do the above problem, with the only difference that the sample size is 16 instead of 81.
In this problem, std. dev. of the mean is 18/4 = 4.5
Using the T-table we find that for 15 dof, P(T>1.753) = 0.05.
So at 1.753 std. dev. above the mean, we will reject H0.
1.753*4.5 = 7.89
So, we will reject H0 at 107.89 or above.
15. Suppose H0: µ = 100 and Ha: µ ≠ 100. Suppose we have a sample of size 81 and a sample standard
deviation of 18 and X-bar of 103?
a. What is the value of the test statistic? (103-100)/(18/sqrt(18)) = 3/2 = 1.5
b. What is the rejection region at alpha of 0.05? >1.96 or <-1.96
c. What will be your decision at alpha of 0.05? Fail to reject H0
d. What is the p-value for this study? 2*P(z>1.5) = 2*(1-0.93319) = 2*0.06681=0.13362
e. Using the p-value, what is your decision? Fail to reject H0
16. Do the above problem, with the only difference that the sample size is 16 instead of 81.
a. What is the value of the test statistic? (103-100)/(18/sqrt(16)) = 3/(4.5) = 0.66667
b. What is the rejection region at alpha of 0.05? >2.132 or <-2.132
c. What will be your decision at alpha of 0.05? Fail to reject H0
d. What is the p-value for this study? You cannot compute this without Excel. Using Excel,
2*P(t>0.66667) = 0.5151
e. Using the p-value, what is your decision? Fail to reject H0
17. Suppose H0: µ = 100 and Ha: µ > 100. Suppose we have a sample of size 81 and a sample standard
deviation of 18 and X-bar of 103?
a. What is the value of the test statistic? (103-100)/(18/sqrt(18)) = 3/2 = 1.5
b. What is the rejection region at alpha of 0.05? >1.645
c. What will be your decision at alpha of 0.05? Fail to reject H0
d. What is the p-value for this study? P(z>1.5) = 1-0.93319 = 0.06681
e. Using the p-value, what is your decision? Fail to reject H0
18. Do the above problem, with the only difference that the sample size is 16 instead of 81.
a. What is the value of the test statistic? (103-100)/(18/sqrt(16)) = 3/(4.5) = 0.66667
b. What is the rejection region at alpha of 0.05? >1.753
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c. What will be your decision at alpha of 0.05? Fail to reject H0
d. What is the p-value for this study? You cannot compute this without Excel. Using Excel,
P(t>0.66667) = 0.2575
e. Using the p-value, what is your decision? Fail to reject H0
19. A small p-value indicates that the actual probability of making a Type-I error is small. T / F
True
20. If we try to reduce the probability of Type-I error, we increase the probability of Type-II error. T / F
True
21. Type-I error is:
a. Rejecting the Null when the Alternate is True.
b. Rejecting the Alternate when the Null is True.
c. Rejecting the Null when the Null is True.
d. Rejecting the Alternate when the Alternate is True.
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