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he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on weekdays and weekends. The owner samples 5 weekdays and finds an average of 259.23 gallons of milk sold on those days with a standard deviation of 34.713. 10 (total) Saturdays and Sundays are sampled and the average number of gallons sold is 365.12 with a standard deviation of 48.297. Construct a 90% confidence interval to estimate the difference of (average number of gallons sold on weekdays - average number of gallons sold on weekends). Assume the population standard deviations are the same for both weekdays and weekends.

Sagot :

Nuhulawal20idn

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;

[tex]S_p^2[/tex] = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

[tex]S_p^2[/tex] = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

[tex]S_p^2[/tex] = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [  4 + 9 ]

[tex]S_p^2[/tex] = [ 4819.9692 + 20993.4 ] / [  13 ]

[tex]S_p^2[/tex] = 25813.3692 / 13

[tex]S_p^2[/tex] = 1985.64378

Now the Standard Error will be;

[tex]S_{x1-x2[/tex] = √[ ( [tex]S_p^2[/tex] / n₁ ) + ( [tex]S_p^2[/tex] / n₂ ) ]

we substitute

[tex]S_{x1-x2[/tex] = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

[tex]S_{x1-x2[/tex] = √[ 397.128756 + 198.564378 ]

[tex]S_{x1-x2[/tex] = √595.693134

[tex]S_{x1-x2[/tex] = 24.4068

Critical Value = [tex]t_{\frac{\alpha }{2}, df[/tex] = [tex]t_{0.05, df=13[/tex] = 1.771  { t-table }

So,

Margin of Error E =  [tex]t_{\frac{\alpha }{2}, df[/tex] × [ ( [tex]S_p^2[/tex] / n₁ ) + ( [tex]S_p^2[/tex] / 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   )

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