Question

1. A large insurance company wants to determine whether the proportion of male policyholders who would not submit auto insurance claims of under $500 is the same as the proportion of female policyholders who do not submit claims of under $500. A random sample of 400 male policyholders produced 272 who had not submitted claims of under $500, whereas a random sample of 300 female policyholders produced 183 who had not submitted claims of under $500

a) Construct a 90% confidence interval for the difference between the proportions of males and of females who had not submitted auto insurance claims of under $500.

b) Find the p-value of the appropriate test.

In a random sample of 10 LAS students, the sample mean time spent studying during a particular week was 15.7 hours with sample standard deviation 3.1 hours. In a random sample of 8 Engineering students, the sample mean time studying during the same week was 20.2 hours per month with sample standard deviation 4.4 hours. Assume that the two populations are normally distributed.

a) Assume that the two population variances are equal. Construct a 95% confidence interval for the difference between the overall average times Engineering and LAS students spent studying during this week.

b) Since the larger sample variance is more than twice as big as the smaller one, the assumption of equal variances is questionable here. Construct a 95% confidence interval for the difference between the overall average times Engineering and LAS students spent studying during this week without assuming that the two population variances are equal. Use Welch’s T.

Answer 1

1)

Test and CI for Two Proportions

Method

p₁: proportion where Sample 1 = Event

p₂: proportion where Sample 2 = Event

Difference: p₁ - p₂

Descriptive Statistics

Sample	N	Event	Sample p
Sample 1	400	272	0.680000
Sample 2	300	183	0.610000

Estimation for Difference

Difference	90% CI for Difference
0.07	(0.009856, 0.130144)

CI based on normal approximation

Test

Null hypothesis	H₀: p₁ - p₂ = 0
Alternative hypothesis	H₁: p₁ - p₂ ≠ 0

Method	Z-Value	P-Value
Normal approximation	1.922	0.0547

The pooled estimate of the proportion (0.65) is used for the tests.

2)

a)

Two-Sample T-Test and CI

Method

μ₁: mean of Sample 1

µ₂: mean of Sample 2

Difference: μ₁ - µ₂

Equal variances are assumed for this analysis.

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Sample 1	10	15.70	3.10	0.98
Sample 2	8	20.20	4.40	1.6

Estimation for Difference

Difference	Pooled StDev	95% CI for Difference
-4.50	3.73	(-8.246, -0.754)

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ ≠ 0

T-Value	DF	P-Value
-2.55	16	0.022

b)

Two-Sample T-Test and CI

Method

μ₁: mean of Sample 1

µ₂: mean of Sample 2

Difference: μ₁ - µ₂

Equal variances are not assumed for this analysis.

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Sample 1	10	15.70	3.10	0.98
Sample 2	8	20.20	4.40	1.6

Estimation for Difference

Difference	95% CI for Difference
-4.50	(-8.506, -0.494)

Test

Null hypothesis	H₀: μ₁ - µ₂ = 0
Alternative hypothesis	H₁: μ₁ - µ₂ ≠ 0

T-Value	DF	P-Value
-2.45	12	0.031