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A bottling plant produces 1 liter bottles of soda. The actual distribution of volumes of soda dispensed to bottles is Normal, with mean μ and standard deviation σ = 0.05 liter. We randomly select 6 bottles andmeasure the volume of soda in each. The results of these 6 measurements (all in liter units) are 1.05 1.04 1.01 1.06 0.94 0.99. Based on these data, a 90% confidence interval for μ is

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shinmin

Given:

Standard deviation = 0.05

Mean = (1.05 + 1.04 + 1.01 + 1.06 + 0.94 + 0.99 / 6) = 1.015

n = 6

Confidence level = 90%

Z score for 90% = 1.645

Solution:

Formula is mean ± z (s / sqrt (n))

= 1.015 ± (1.645) (0.05/ sqrt (6)

= 1.015 ± (1.645) (0.020412414)

= 1.015 ± 0.033578421

= 0.9814 < x < 1.0486 is the confidence interval

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