A summer intern at the Lego factory in Billund, Denmark, samples 42 yellow brick

Question

A summer intern at the Lego factory in Billund, Denmark, samples 42 yellow bricks and 26 red bricks and measures their weight. She finds sample means of 4.1 grams (yellow bricks) and 4.2 grams (red bricks), and sample standard deviations of 0.15 (yellow) and 0.23 (red), respectively. (a) Compute the 95% confidence interval for the mean weight of red bricks. (b) Compute the 95% confidence interval for the difference between the mean weights (i.e., red minus yellow). (c) Compute the 95% confidence interval for the variance of the weight for the yellow bricks. For each question above, make sure you state clearly which distribution you are using, what assumptions that entails, and whether there could be alternate choices.

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=4.2
Standard deviation( sd )=0.23
Sample Size(n)=26
Confidence Interval = [ 4.2 ± t a/2 ( 0.23/ Sqrt ( 26) ) ]
= [ 4.2 - 2.06 * (0.045) , 4.2 + 2.06 * (0.045) ]
= [ 4.107,4.293 ]
b.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=4.1
Standard deviation( sd1 )=0.15
Sample Size(n1)=42
Mean(x2)=4.2
Standard deviation( sd2 )=0.23
Sample Size(n2)=26
CI = [ ( 4.1-4.2) ±t a/2 * Sqrt( 0.0225/42+0.0529/26)]
= [ (-0.1) ± t a/2 * Sqrt( 0.0026) ]
= [ (-0.1) ± 2.06 * Sqrt( 0.0026) ]
= [-0.2044 , 0.0044]
c.
Confidence Interval
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S^2 = Variance
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size
Since aplha =0.05
^2 right = (1 - Confidence Level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 41 df are 60.5606 , 25.215
Variacne( S^2 )=8.4
Sample Size(n)=42
Confidence Interval = [ 41 * 8.4/60.5606 < ^2 < 41 * 8.4/25.215 ]
= [ 344.4/60.5606 < ^2 < 344.4/25.2145 ]
[ 5.6869 , 13.6588 ]

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