Final Exam Ec315

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PART I. HYPOTHESIS TESTING
PROBLEM 1 A certain brand of fluorescent light tube was advertised as having an effective life span before burning out of 4000 hours. A random sample of 84 bulbs was burned out with a mean illumination life span of 1870 hours and with a sample standard deviation of 90 hours. Construct a 95 confidence interval based on this sample and be sure to

interpret this interval.

Answer
Since population standard deviation is unknown, t distribution can be used construct the confidence interval.



The 95% confidence interval is given by  X  t / 2,n 1



S
S
, X  t /2,n 1

n
n

Details
Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation
Sample Mean
Sample Size
Confidence Level

90
1870
84
95%

Intermediate Calculations
Standard Error of the Mean
9.819805061
Degrees of Freedom
83
t Value
1.988959743
Interval Half Width
19.53119695
Confidence Interval
Interval Lower Limit
1850.47
Interval Upper Limit
1889.53

2
PROBLEM 2 Given the following data from two independent data sets, conduct a one -tail hypothesis test to determine if the means are statistically equal using alpha=0.05. Do NOT do a confidence interval.

n1 = 35
n2 = 30
xbar1= 32
xbar2 = 25
s1=7
s2 = 6
Answer
H0:µ1=µ2
H1: µ1>µ2
Test statistics used is t 

X1  X 2
S

2
(n1  1) S12  (n2  1) S2
n1n2
~ tn1  n1 2 where S 
n1  n2  2
n1  n2

Decision rule : Reject the null hypothesis, if the calculated value of test statistic is greater than the critical value.
Details
t Test for Differences in Two Means
Data
Hypothesized Difference
Level of Significance
Population 1 Sample
Sample Size
Sample Mean
Sample Standard Deviation
Population 2 Sample
Sample Size
Sample Mean
Sample Standard Deviation

0
0.05
35
32
7
30
25
6

Intermediate Calculations
Population 1 Sample Degrees of Freedom
34
Population 2 Sample Degrees of Freedom
29
Total Degrees of Freedom
63
Pooled Variance
43.01587
Difference in Sample Means
7
t Test Statistic
4.289648
Upper-Tail Test
Upper Critical Value
p-Value
Reject the null hypothesis

1.669402
3.14E-05

Conclusion: Reject the null hypothesis. The sample provides enough evidence to support the claim that means are different.

3
PROBLEM 3. A test was conducted to determine whether gender of a display model af fected the likelihood that consumers would prefer a new product. A survey of consumers at a trade show which used a female spokesperson determined that 120 of 300 customers preferred the product while 92 of 280 customers preferred the product when it was shown by a female spokesperson.

Do the samples provide sufficient evidence to indicate that the gender of the salesperson affect the likelihood of the product being favorably regarded by consumers? Evaluate with a two-tail, alpha =.01 test. Do NOT do a confidence interval.

Answer
H0: There no significant gender wise difference in the proportion customers who preferred the product.
H1: There significant gender wise difference in the proportion customers who preferred the product.
P  P2
n p n p
1
The test Statistic used is Z test Z 
where p= 1 1 2 2
n1  n2
1 1
P(1  P)   
 n1 n2 
Decision rule : Reject the null hypothesis, if the calculated value of test statistic is greater than the critical value.
Details
Z Test for Differences in Two Proportions
Data
Hypothesized Difference
Level of Significance
Group 1
Number of Successes
Sample Size
Group 2
Number of Successes
Sample Size

0
0.01
Male
120
300
Female
92
280

Intermediate Calculations
Group 1 Proportion
0.4
Group 2 Proportion
0.328571429
Difference in Two Proportions 0.071428571
Average Proportion
0.365517241
Z Test Statistic
1.784981685
Two-Tail Test
Lower Critical Value
-2.575829304
Upper Critical Value
2.575829304
p-Value
0.074264288
Do not reject the null hypothesis

Conclusion: Fails to reject the null...
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