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BUSA 3321 - EXAM 2 - SUMMER, 1997 - Version White, Name
_______________________
Questions 1-6 are worth 5 points each.
1. A simple random sample is a requirement of most of the inferences in a first statistics
course. What is the importance of a simple random sample?
2 Define "sampling distribution".
3. The central limit theorem states that as the sample size gets larger....
4. What is the meaning of a type I error?
5. Complete the following sentence. In a 90% confidence interval for a mean, 90% of the
time ......
6. An investor is interested in the average sales of companies in Ohio. To estimate this
value, the investor samples 16 companies in various cities in Ohio and records their sales.
For the t distribution, the population needs to be normal. Complete the following sentence.
In this situation, the population of
_____________________________________________________ needs to be normal.
7. You want to determine if the mean of a distribution exceeds $3.00. A sample of size 81
produces a sample mean of 3.1 and  = 0.9. What is the test statistic value?
8. What will be a 95% confidence interval for a mean of a normal distribution given a
sample mean = 3.1, n = 49, and  = 0.7? No conclusion necessary.
9. Given a sample proportion of .4 and a sample size of 100, what is the test statistic value
for testing
H1: P > .25?
10. Given that you are doing a large sample test of the mean (H1:  > 20), you have that the
sample mean is 25, the sample size is 100 and the population standard deviation is 37.037?
What is the p-value? Show graph and indicate shaded region.
11. The weekly cost for students at Kiddie Care Day Care Center is normally distributed
with a mean of $50 and a standard deviation of $20. A sample of size 25 is taken. 80%
percent of the sample means will be more than _____ dollars?
graph
answer (show work for partial credit)
12. Given  = $10, find the sample size necessary to estimate the mean of a normal distribution
within $0.10 with 99% confidence.
Formula:
Substitution:
Value:
1
3. A television producer is interested in the percent of viewers that are watching a specific
show. Last month only eight percent of the television audience was estimated to be
watching. How many viewers would have to be interviewed to estimate the percentage this
month? You want the estimate to be within one percentage point of the true percent with
99% confidence.
Formula:
Substitution:
Value:
14. Find the following p-values (if the probability is between two values in the table then
indicate this as a range; e.g., prob = 0.05 to 0.025)
a. H1  > 20 and z* = 2.55
b. H1  < 20 and t* = 2.32 with n = 8
c. H1 P > .20 and z* = 2.38
15. A physician is concerned with the average monthly expenditure on liability insurance.
She heard that the average is greater than $1000. She plans to take a random sample of 100
physicians to see if this claim is true. Assume the population standard deviation is known.
What would be the null and alternative hypothesis and the rejection region ( = 0.01)? Do
not calculate a test statistic or make a conclusion. Asssume  is known.
Ho:
R.R. Reject Ho if
H1:
16. Suppose a physician is testing the alternative hypothesis that the average monthly
expenditure on liability insurance is greater than $1000. A sample of size 64 yields a z-test
value of 2.33. Given = 0.05, what conclusion should she make? Give all parts to the
conclusion.
17. A real estate agent wants to know if the average price of houses that are being sold in a
city is less than $100,000. A random sample of 25 recently sold houses revealed an average
house price of $90,000 with a standard deviation of $30,000. Is there evidence to support
the agent's claim? For this problem only give the Null and Alternative hypothesis along
with the rejection region. Do not calculate a test statistic or make a conclusion.
Ho:
H1:
R.R. Reject Ho if
18. Another real estate agent also wants to if know the average price of houses that are
being sold in their city is less than $100,000. A sample of size 16 houses results in a p-value
of 0.03. Is there evidence to support the agent's claim? For this problem indicate the
managerial conclusion (giving all parts) and state any necessary assumption(s).
Conclusion
Assumption(s)
19. A discount store claims that its steel belted radial is more resistant to wear than it
competitors. On 40 cars, one tire from the discount store and one from the competition was
mounted on the rear axle. After each car was driven 8,000 miles, the tire with the less wear
was recorded. Suppose the tire of discount store showed less wear on 21 of the cars. What
would you conclude about the discount store's claim? Use  = 0.05.
Ho:
R.R. Reject Ho if z > 1.645
H1:
Test Statistic Z value =
Con: At  = 0.05 we (can / can not ) say that
_________________________________________________________________________
____________________________________________________________
20. An instructor wonders if the scores in the last exam were typical of previous exams. In
the past the score of the exam averaged 75. In this class of 9 students, the class average was
80 with a standard deviation of 12. Is there enough evidence to say that this exam was too
easy? Use  = 0.05.
Ho:
H1:
Test Statistic =
R.R. Reject Ho if t ( 8, 0.05) > 1.860
Con: At a = 0.05 we (can / can not ) say that_________________________
_________________________________________________________________________
____________________________
ANSWERS BUSA 3321 - EXAM 2 - SUMMER, 1997 - Version White
Questions 1-6 are worth 5 points each.
1. Random Sample: Each sample of size n (and therefore every observation) has the same
chance of being selected. The chance that the mean takes on certin values can be computed
from this.
2 "sampling distribution". Taking different samples results in different values of a
statistic (e.g., the mean). The distribution of these values is called the sampling distriubtion
of the statistic.
3. Ccentral limit theorem the distribution of the sample mean approaches a normal
distribution regardless of the distribution of the origninal values.
4. type I error: Reject a true null hypothesis.
5. In a 90% confidence interval for a mean, 90% of the time ......the interval calculated
in the sample will contain the population mean.
6. An investor is interested in the average sales of companies in Ohio. In this situation,
the population of sales of all companines in Ohio needs to be normal.
7. mean exceeds $3.00. n= 81, a sample mean of 3.1 and s = 0.9. What is the test
statistic value?
Formula: (x - hypothesized value ) / [  / n ] = (3.1 - 3) / [.9 / 81] = 1
8. 95% C.I. mean, sample mean = 3.1, n = 49, and  = 0.7?
Formula: x  Z /2 [ / n] = 3.1  1.96 [0.7 / 49] = 2.904 to 3.296
9. phat = .4 and n=100, what is the test statistic value for testing H1: P > .25?
pˆ  p
0.40  .25

 3.46
Formula:
p(1  p)
0.25(0.75)
n
100
10. (H1:  > 20), x = 25, n=100,  = 37.037? What is the p-value?
z = (x - ) / [ / n] = ( 25 - 20 ) / [37.037 / 100] = 1.35; p-value = Pr (Z > 1.35) =
.5 - .4115 = .0885
11.  = $50 and  = $20. n= 25. Pr ( x > ? ) = .80
Since 80% of the values are larger than this number, the value must be to the left of .
There is an area of 30% between and the value and . Look up 0.3000 and you will find a
Z-value of 0.84. Since this is to the left of the mean, the z-value is –0.84. Substitute the
values for  , , n and Z into the formula below and solve for xbar.
z = (x - ) / [ / n] or x = z [ / n] +  = [-.84]* 20 / 25 + 50 = $46.64
graph:
12. Given  = $10, e = $0.10 with 99% confidence.
n = z22 / e2 = (2.575)2 102 / (.10)2 = 66,307
13. e = .01, 99% confidence, use p =.08.
n = z2p(1-p) / e2 = (2.575)2 0.08*0.92 / (.01)2 = 4881
14. a. H1  > 20 and z* = 2.55
Pr(Z>2.55) = 1 - 0.9946 = 0.0054
b. H1  < 20 and t* = 2.32 with n = 12 Pr (t < 2.32) = 1 - Pr(t>2.32) = 1 - (.025 to .01) =
.975 to .99
c. H1 p > .20 and z* = 2.38
Pr (Z > 2.38) = 1 - .9913 = 0.0087
15. the average is greater than $1000. n= 100 ( = 0.01)?
Ho:  = 1000
H1:  > 1000
R.R. Reject Ho if Z > 2.33
16. average monthly expenditure on liability insurance is greater than $1000. n=64;
z= 2.33. = 0.05,
Reject Ho if Z > 1.645, therefore
At  = 0.05 we can say that average monthly expenditure on liability insurance is greater
than $1000
17. ...the average price of houses ....is less than $100,000; n= 25 recently  unknown
Ho:  = 100,000
H1:  < 100,000
R.R. Reject Ho if t < t 24, 0.05
18. ...the average price of houses that are being sold in their city is less than $100,000.
n=16; p-value of 0.03.
Conclusion At a = 0.05 we can say that the average price of houses that are being sold in
their city is less than $100,000
Assumption(s) You have a random sample of houses and the price of houses that are being
sold in their city is normally distributed.
19. 19. measuring success (discount store's tires are more resistant to wear than its
competitors) or failure (discount store's tires are not more resistant to wear than its
competitors) Out of n = 40 we find 21 successes. Ps = 21/40 . If discount store is more
resistant then P > .50
Ho: p = .50
H1: p > .50
T.S Formula and substitution::
R.R. Reject Ho if z > 1.645
Z
pˆ  p
0.525  0.50

 0.316
p(1  p)
0.50(0.50)
n
40
Con: At  = 0.05 we can not say that the percent of time that the discount store is
more resistant than the competitors exceeds 50%. We can not support the discount
store's claim.
20. the exam averaged 75. n=9, x=80 s=12. If exam was too easy, then  > 75.  = 0.05.
Ho:  = 75
1.860
R.R. Reject Ho if t ( 8, 0.05) >
H1:  > 75
T.S Formula and substitution::
( x -  )
80 - 75
t = ------------------ = ---------------- = 1.25
s / n
12 / 9
Con: At  = 0.05 we can not say that the mean of the current exam (if given to all
students) exceeds 75. We can not say that the exam was too easy.