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...