Each of the 492 people who came to a school football game paid $12 for a ticket. How much ticket money did the school collect?
2 answers:
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Answer:
90% confidence interval is ( -149.114, -62.666 )
Step-by-step explanation:
Given the data in the question;
Sample 1 Sample 2
x"₁ = 259.23 x"₂ = 365.12
s₁ = 34.713 s₂ = 48.297
n₁ = 5 n₂ = 10
With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }
significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1
Since we are to assume that variance are equal and they are know, we will use pooled variance;
Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13
Now, pooled estimate of variance will be;
= [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]
we substitute
= [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]
= [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [ 4 + 9 ]
= [ 4819.9692 + 20993.4 ] / [ 13 ]
= 25813.3692 / 13
= 1985.64378
Now the Standard Error will be;
= √[ (
/ n₁ ) + (
/ n₂ ) ]
we substitute
= √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]
= √[ 397.128756 + 198.564378 ]
= √595.693134
= 24.4068
Critical Value = =
= 1.771 { t-table }
So,
Margin of Error E = × [ (
/ n₁ ) + (
/ n₂ ) ]
we substitute
Margin of Error E = 1.771 × 24.4068
Margin of Error E = 43.224
Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89
So, Limits of 90% CI will be; x₁ - x₂ ± E
Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114
Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666
Therefore, 90% confidence interval is ( -149.114, -62.666 )