Download Spring 2005 exam 2 solutions

BUSA 3321 - EXAM 2 – Spring, 2005 - Version White
1. What will be a 90% confidence interval for a mean of a normal distribution given x = 36, n = 81, and
 = 18?
Since  is known use a Z confidence interval A 90% interval uses a Z of 1.645
X Z

n
 36  1.645
18
 36  3.29
81
2. A medical test indicates that you have a disease. The status quo is that you do not have the disease.
What is the consequence of a Type 1 error in this case?
Null = you do not have the disease. A type 1 error is rejecting the null when it is true. That is the
test says you are sick when actually you are not, false positive. You would be getting treatments
that you do not need
3. What t-table value would use if n=35 and confidence is 0.99?
go to row 34 and column 0.005, The t value is 2.7284
4. A professor wishes to determine how difficult an exam is by trying it on a few students. From a random
sample of 16 students you find an average grade of 64. If you assume a normal distribution of grades,
what is your estimate of the mean grade if the test is applied to all students? (assume  = 16)
Since  is known use a Z confidence interval. The default confidence gives a Z value of 1.96
X Z

n
 64  1.96
16
 64  7.84 . None of the above
16
5. What is the probability that Z < -2.45 ?
Pr( Z < -2.45) = 0.0071
6. What would be the margin of error for a 90% confidence interval for a mean if n=25 and =8?
Margin of error for a mean with  know is = Z

n
 1.645
8
25
 2.632
7. Given P=0.50 and n = 25. Find Pr(Ps > 0.42 ).
Z
Ps  P
P(1  P)
n

0.42  0.50
 0.8
0.5(1  0.5)
25
Pr ( Z > -0.8) = 0.5000+.2881 = 0.7881
8. Which of the following is required for all inferences in chapter 6-8?
Some inferences require a normal distribution or a large sample size. Some require a sample size
such that np and n(1-p) are greater than five but all require a simple random sample.
9. If you wish to estimate the percentage of cars that drive faster than the speed limit to within  4%, how
many observations would you need?
This is a question asking about the sample size for a proportion. The formula is n 
Z 2 p(1  p)
e2
For 95% confidence (the default level) z = 1.96. Since p is unknown we use 0.5 and the margin of
error is 0.04.
Z 2 p(1  p) (1.96) 2 0.5(1  0.5)
n

 600.25
e2
(0.04) 2
Rounding up the sample size needed is 601
10 You are trying to estimate the average speed of cars on I-20. Based on your random sample you are
90% confident that the average speed was 70  15 miles per hour. If you wish to reduce the margin of
error to 5 miles per hour what would you have to do?
increase your sample size and/or decrease your confidence
11. You are measuring the average weight of catsup in catsup bottles ( known). You are concerned that
the true average is becoming smaller than the labeled weight. You have the information from a random
sample of 36 bottles. What would be the rejection region ( = 0.10)? Reject Ho if
Since  is known and you must use a Z rejection region. (additionally the population should
be normally distributed)
This is a one-sided test to the left, since you are concerning that the average has become
smaller.
the rejection region is Z < -1.28.
12. Which of the following is required before you can use a standard normal with a binomial proportion?
From your notes the requirements for a binomial proportion to be approximated by a Z is
i. simple random sample
ii. sample size large enough that np and n(1-p) are both greater than 5.
The answer then is none of the above
13. If the average error in predicting temperature in the state based on a random sample of 25 cities is
1.5 degrees  0.25, what conclusion can you make?
A. With 95% confidence the average prediction error in these 25 cities is 1.5 degrees with a margin of
error of  0.25 degrees
B. With 95% confidence the average prediction error in all cities is 1.5 degrees with a margin of error
of  0.25 degrees
Since you are 100% sure that the sample mean is 1.5, the answer can not be A. You are taking a
random sample from all cities, therefore your inference is about all cities.
14. next page
15. Your company manufactures batteries that have an average life of 48 months with a standard
deviation of 2.5 months. 90% of the batteries will last at least how long (only 10% of the batteries will
last longer)?
You want to find the value such that Pr (X < ? ) = 0.90. This X values corresponds to a Z value of
1.28. The X value then is 1.28 standard deviations above the mean = 48 + 1.28 (2.5) = 51.2, none of
the above
16. A company produces ball bearings that are on average 50 millimeters in diameter with a standard
deviation of 0.1 millimeters. Any bearing more than 0.25 millimeters away from the average has to be
scrapped. What percent have to be scrapped?
0.25 millimeters is 2.5 standard deviations (0.25/0.1). You wish to determine the chance of finding
a value more than 2.5 standard deviations from the mean (to the left or to the right). The
probability of being less than -2.5 is 0.0062. Therefore the chance of being beyond is 00.124
.
14. If 99% of the time the grades in a class fall between 55 and 95, what is the variability of the grades?
Assume grades have a normal distribution
17. Suppose you are testing the null hypothesis that the average return on investment for stocks in
England is greater than or equal to 39 pounds. A sample of size 81 stocks a z-test value of –1.67.
What decision do you make about the null hypothesis?
The null is that the mean is greater than or equal to 39. The rejection region is then to the left.
The p-value is then the Pr( Z < -1.67) = 0.0475. Since this is smaller than 0.05 (the default  ) we can
reject the null hypothesis.
18. You want to determine if the mean cost of a cake in the state differs from $5.00. You are given  =
1.8 from a normal distribution. From a random sample of size 81 you calculate a test statistic value of
Z=2.98. What conclusion can you make?
Since the equal sign must be in the null. The null is that the mean equals $5,
The rejection region is a two sided z region. Therefore reject the null if the Z > 1.96 or Z < -1.96.
Since 2.98 > 1.96 reject the null and conclude the mean of the cake differs from 5.
19. What will be a 95% confidence interval for a population proportion given Ps = 0.2 and n = 1600
Ps  Z
Ps (1  Ps )
0.2(1  0.2)
 0.20  1.96
 0.2  0.0196
n
1600
BUSA 3321 - EXAM 2 – Spring, 2005 - Version Blue,
1. What is the probability that Z > -2.45 ?
Pr( Z > -2.45) = 1 – Pr(Z < -2.45) = 1 – 0.0071 = 0.9929
2. Given P=0.50 and n = 25. Find Pr(Ps > 0.42 ).
Z
Ps  P
P(1  P)
n

0.42  0.50
 0.8
0.5(1  0.5)
25
Pr ( Z > -0.8) = 0.5000+.2881 = 0.7881
3. A medical test indicates that you have a disease. The status quo is that you do not have the disease.
What is the consequence of a Type 2 error in this case?
Null = you do not have the disease. A type 2 error is accepting the null when it is false. That is the
test says you are not sick when actually you are, false negative. You would not be getting treatments
that you do need.
4. What would be the margin of error for a 90% confidence interval for a mean if n=25 and =8?
Margin of error for a mean with  know is = Z

n
 1.645
8
25
 2.632
5. A company produces ball bearings that are on average 50 millimeters in diameter with a standard
deviation of 0.1 millimeters. Any bearing more than 0.25 millimeters away from the average has to be
scrapped. What percent have to be scrapped?
0.25 millimeters is 2.5 standard deviations (0.25/0.1). You wish to determine the chance of finding
a value more than 2.5 standard deviations from the mean (to the left or to the right). The
probability of being less than -2.5 is 0.0062. Therefore the chance of being beyond is 0.0124
6. Which of the following is required before you can use a standard normal with a binomial proportion?
From your notes the requirements for a binomial proportion to be approximated by a Z is
i. simple random sample
ii. sample size large enough that np and n(1-p) are both greater than 5.
The answer then is none of the above
7. You want to determine if the mean cost of a cake in the state differs from $5.00. You are given  = 1.8
from a normal distribution. From a random sample of size 81 you calculate a test statistic value of
Z=2.98. What conclusion can you make?
Since the equal sign must be in the null. The null is that the mean equals $5,
The rejection region is a two sided z region. Therefore reject the null if the Z > 1.96 or Z < -1.96.
Since 2.98 > 1.96 reject the null and conclude the mean of the cake differs from 5.
8. What will be a 90% confidence interval for a mean of a normal distribution given x = 36, n = 81, and
s = 18?
Since  is unknown use a t confidence interval with 80 degrees of freedom. A 90% interval uses a
t from column 0.05 and row 80 of the t-table
X t
S
n
 36  1.6641
18
81
 36  3.33
9. What will be a 95% confidence interval for a population proportion given Ps = 0.2 and n = 1600
Ps  Z
Ps (1  Ps )
0.2(1  0.2)
 0.20  1.96
 0.2  0.0196
n
1600
10. You are trying to estimate the average speed of cars on I-20. Based on your random sample you are
90% confident that the average speed was 70  15 miles per hour. If you wish to reduce the margin of
error to 5 miles per hour what would you have to do?
increase your sample size and/or decrease your confidence
11. Your company manufactures batteries that have an average life of 48 months with a standard
deviation of 2.5 months. 90% of the batteries will last at least how long (only 10% of the batteries will
last longer)?
You want to find the value such that Pr (X < ? ) = 0.90. This X values corresponds to a Z value of
1.28. The X value then is 1.28 standard deviations above the mean = 48 + 1.28 (2.5) = 51.2, none of
the above
12. If you wish to estimate the percentage of cars that drive faster than the speed limit to within  4%,
how many observations would you need?
This is a question asking about the sample size for a proportion. The formula is n 
Z 2 p(1  p)
e2
For 95% confidence (the default level) z = 1.96. Since p is unknown we use 0.5 and the margin of
error is 0.04.
Z 2 p(1  p) (1.96) 2 0.5(1  0.5)
n

 600.25
e2
(0.04) 2
Rounding up the sample size needed is 601
13. A professor wishes to determine how difficult an exam is by trying it on a few students. From a
random sample of 16 students you find an average grade of 64. If you assume a normal distribution of
grades, what is your estimate of the mean grade if the test is applied to all students? (assume  = 16)
Since  is known use a Z confidence interval. The default confidence gives a Z value of 1.96
X Z

n
 64  1.96
16
 64  7.84 . None of the above
64
14. Which of the following is required for all inferences in chapter 6-8?
Some inferences require a normal distribution or a large sample size. Some require a sample size
such that np and n(1-p) are greater than five but all require a simple random sample.
15. What t-table value would use if n=35 and confidence is 0.99?
go to row 34 and column 0.005, The t value is 2.7284
16. You are measuring the average weight of catsup in catsup bottles. You are concerned that the true
average is becoming smaller than the labeled weight. All the information you will have comes from a
random sample of 36 bottles. What would be the rejection region ( = 0.10)? Reject Ho if
Since the only information you have comes from a sample,  is unknown and you must use a t
rejection region. (additionally the population should be normally distributed)
This is a one-sided test to the left, since you are concerning that the average has become
smaller.
Go to row 35 and column 0.10, the rejection region is t < -1.3062.
17. Suppose you are testing the null hypothesis that the average return on investment for stocks in
England is greater than or equal to 39 pounds. A sample of size 81 stocks a z-test value of –1.67.
What decision do you make about the null hypothesis?
The null is that the mean is greater than or equal to 39. The rejection region is then to the left.
The p-value is then the Pr( Z < -1.67) = 0.0475. Since this is smaller than 0.05 (the default  ) we can
reject the null hypothesis.
18. If the average error in predicting temperature in the state based on a random sample of 25 cities is 1.5
degrees  0.25, what conclusion can you make?
a. With 95% confidence the average prediction error in all cities is 1.5 degrees with a margin of error
of  0.25 degrees
b. With 95% confidence the average prediction error in these 25 cities is 1.5 degrees with a margin of
error of  0.25 degrees
Since you are 100% sure that the sample mean is 1.5, the answer can not be b. You are taking a
random sample from all cities, therefore your inference is about all cities.
19. If 99% of the time the grades in a class fall between 55 and 95, what is the variability of the grades?
Assume grades have a normal distribution