101k views
3 votes
The following information was obtained from independent random samples taken of two populations. Assume normally distributed populations with equal variances. Sample 1 Sample 2 Sample Mean 45 42 Sample Variance 85 90 Sample Size 10 12 The 95% confidence interval for the difference between the two population means is (use rounded standard error) a. -2.65 to 8.65. b. -5.344 to 11.344. c. -5 to 3. d. -4.86 to 10.86.

User Hmallett
by
8.1k points

1 Answer

2 votes

Answer:

95% Confidence Interval is; ( -5.344 to 11.344 )

Option b) -5.344 to 11.344 is the correct answer

Explanation:

Given the data in the question;

Sample 1 Sample 2

x"₁ = 45 x"₂ = 42

S₁² = 85 S₂² = 90

n₁ = 10 n₂ = 12

df = [ S₁²/n₁ + S₂²/n₂ ]² / [ ((S₁²/n₁)²/n₁-1) + ((S₂²/n₂)²/n₂-1)) ]

we substitute

df = [ 10/10 + 90/12 ]² / [ ((85/10)²/10-1) + ((90/12)²/12-1)) ] = 19.64 ≈ 20

df = 20

with 95% confidence interval

∝ = 1 - 0.95 = 0.05

∝/2 = 0.05/2 = 0.025

now,
t_(\alpha /2,df) = t_(0.025, 20) = 2.086 { from table }

95% confidence interval for N1 - N2

⇒ (x"₁ - x"₂) ±
t_(\alpha /2,df) × √( S₁²/n₁ + S₂²/n₂ )

⇒ (45 - 42) ± 2.086 × √( 85/10 + 90/12 )

⇒ 3 ± 2.086 × 4

⇒ 3 ± 8.344

so;

Lower Limit = 3 - 8.344 = -5.344

Upper Limit = 3 + 8.344 = 11.344

Therefore, 95% Confidence Interval is; ( -5.344 to 11.344 )

Option b) -5.344 to 11.344 is the correct answer

User Pashaplus
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories