Download 1 - Academic Information System (KFUPM AISYS)

Quiz #6DE
Semester 061
Dept of Mathematical Sciences
Stat319: Intro to Prob & Stat for Engineers & Scientists Instructor: Musawar A. Malik
Section: 2 (8-8.50am), 4 (9-9.50am)
Name:_______________________ ID#:______________ Serial #:_____________________
Directions: This quiz is designed for 40 minutes. Please answer all questions. Please show your steps and box
your final answers for the following problem solving questions.
1.
A random sample of n = 500 people were surveyed recently to determine an estimate for the proportion of
people in the population who had attended at least some college. With a sample proportion of 0.40, the survey
estimated that between 0.357 and 0.443 of the population had attended. Given this information,
a. What is the point estimate?
b. What is the margin of error?
c. What is the standard error?
d. What approximately is the confidence level for the interval estimate constructed in the survey?
Answer: [5pts total]
a) [1pt] Point estimate = middle of interval =(LCL+UCL)/2 = (0.357+0.443)/2=0.8/2=0.4
b) [1pt] margin of error = UCL – point estimate = 0.443 - 0.4 = 0.043
c) [1pts] standard error = (pq/n) = square root of (0.4*0.6/500) = (0.00048) =0.0219089
d) [2pts] 0.443 = 0.4 + z/2*sp  0.443 = 0.4 + z/2*(0.0219089)
 z/2 =(0.443 - 0.4)/(0.0219089) =1.96267  1.96. So  = 0.05 from the z-table. And confidence level = 1-  = 1- 0.05 = 0.95
Level (Moderately Difficult)
Keywords: population proportion, confidence level, sample size
2.
ARAMCO human resources manager wishes to estimate the proportion of employees who are
engineers in the company. What is the minimum size sample he should select if he wants 95 percent
confidence and a margin of error less than + 0.01?
Solution: [3pts total ]
He is interested in estimating
 , the population proportion.
The sample size formula is:
A conservatively large sample size can be developed by substituting 0.50 for
n
1.962 (0.50)(1  0.50)
0.012
.
n
z 2 (1   )
.
e2
This gives:
 9, 604 . {Z=1.96 [1pt], e =0.01 [1pt], rounded final answer[1pt]}
Moderate
Keywords: sample size, population proportion, confidence, margin of error
3.
A PC company uses two suppliers for rechargeable batteries for their notebook computers. Two
factors are important quality features of the batteries: mean time of use and variation. It is desirable
that the mean use time is high and the variability is low. Recently, the PC maker conducted a test on
batteries from the two suppliers to decide whether they are different. In the test, 9 randomly selected
batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested.
The following results were observed:
Supplier 1
n1 = 9
Supplier 2
n2 = 12
x1  67.25 min.
x2  72.4 min.
s1 = 11.2 min
s2 = 9.9 min.
Based on these sample results,
a) What is the point estimate for the difference in the two batteries?
b) Construct a 90% confidence interval estimate for the difference between the two batteries with respect to
the population mean use time? [Hint: you can assume equal population variance]
c) With the information constructed in (b), can you conclude that the two batteries are probably different?
ANSWER: [7pts total]
a) [1pt] point estimate =(67.25-72.4) = -5.15
b) [5pts total] The t critical value is found from the t-distribution with degrees of freedom equal to
n1 + n2 – 2 = 9 + 12 – 2 = 19. [1pt]
Given a 0.10 significance level, the critical value from the t-distribution with degrees of freedom equal to 19 is equal
to +1.729. [1pt]
Quiz #6DE
Semester 061
Dept of Mathematical Sciences
Stat319: Intro to Prob & Stat for Engineers & Scientists Instructor: Musawar A. Malik
Section: 2 (8-8.50am), 4 (9-9.50am)
Name:_______________________ ID#:______________ Serial #:_____________________
Directions: This quiz is designed for 40 minutes. Please answer all questions. Please show your steps and box
your final answers for the following problem solving questions.
sp is the pooled standard deviation and is computed as:
(n1  1)s12  (n 2  1)s 22
(8)11.22  (11)9.92

 10.4671. [1pt]
n1  n 2  2
19
sp 
Then the 90% Confidence interval estimate for the population mean difference is:
( x 1  x 2 )  t  / 2s p
1 1
1 1

= (67.25-72.4) + 1.729(10.4671)

n1 n 2
9 12
[1pt]
= -5.15 + 1.729(10.4671)(0.440959)
= -5.15 + 7.9803
= ( -13.4803 , 2.4803)
[1pt]
c) [1pt] Since 0 (1- 2 = 0) is in the interval constructed in (b), we do not reject the null hypothesis. Thus, based on the
sample data, we do not have evidence to indicate that the two batteries differ in mean use time.
Moderate
Keywords: t-test, mean difference, independent
4.
A university computer lab manager wishes to estimate the mean time that students stay in the lab per
visit. He believes that the population standard deviation would be 10 minutes.
a. He further wishes to have a confidence level of 90 percent and a margin of error of + 2.00 minutes.
Find the required sample size needed.
b. A random sample of size 64 was obtained with a mean of 18. Construct a 90% confidence interval for
the population mean.
c. Using a significance level of 0.01 and the sample found in (b), test the hypothesis that students typically
spend at most 15.5 minutes in the lab per visit.
d. Find the p-value for the test in (c).
e. What is your decision in (c) if a significance level of 0.05 was used instead?
Solution: [15pts total ]
a) [2pts] He is interested in estimating , the population mean. Since we are given , substituting required information into the
formula gives:
n
1.6452 (102 )
z 2 2
n

 67.6506 . Given this, the sample size is approx 68 students.

e2
22
b) [5pts total] Given a 0.10 significance level, the critical value from the z-distribution is equal to 1.645. [1pt]
Then the 90% Confidence interval estimate for the population mean difference is:
(x )  z  / 2
c)

n
= 18 + 1.645
10
= 18 + 1.645*10/8
64
[3pts]
= 18 + 2.05625 = ( 15.9438 , 20.05625)
[1pt]
[5pts total] H0:  < 15.5 versus Ha:  > 15.5. [2pts]
Given a 0.01 significance level, the critical value for an upper 1-tailed test from the z-distribution equal to 2.33. [1pt]
So, the critical region is Z > 2.33. The z-test statistic is computed as:
Z 
x   18  15.5

 2.0 . [2pts]
 n 10 / 64
Do Not Reject Ho. [1pt] Insufficient evidence to conclude that students spend more than 15.5 minutes at the lab. [1pt]
d) [1pt] P-value = P( Z > 2.0) = 0.0228 from the z-table.
e) [2pts] H0:  < 15.5 versus Ha:  > 15.5.
Given a 0.05 significance level, the critical value for a 1-upper tailed from the z-distribution equal to 1.645. [1pt] So, the
critical region is Z > 1.645. The z-test statistic is computed as:
Z 
x   18  15.5

 2.0 .
 n 10 / 64
Reject Ho. Enough evidence to conclude that students spend more than 15.5 minutes at the lab. [1pt].
Or P-value (0.0228) <  (0.05), so reject Ho.
Level (Moderate)
Keywords: sample size, estimate, confidence level