Ounce Lab Balance
A look at what is currently available on eBay
 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $10.98 
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 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $14.99
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 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $20.97
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 100 x 0.01 Gram Electronic Digital Lab Balance Scale oz US $2.89
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 Electronic Digital Lab Balance Scale oz 500 x 0.1 Gram US $6.39
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 Sweet Spot Labs Balancing Mist Aroma 11 4 oz / 120 ml US $15.00
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 BEHAVIOR BALANCE-DMG LIQUID by DAVINCI LABS 12oz AUTISM US $24.95
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 Behavior Balance-DMG Liquid 12 oz by Davinci Labs US $27.53
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 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $14.99
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 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $10.98
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 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $20.97
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 600 X 0.1 GRAM DIGITAL SCALE 0.1 Ounce Lab Balance New US $14.99
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Confidence Iinterval problem. Statistics. I need help!?
Suppose your class is investigating the weights of Snickers 1-ounce Fun-Size candy bars to see if customers are getting full value for their money. Assume that the weights are Normally distributed with standard deviation = 0.005 ounces. Several candy bars are randomly selected and weighed with sensitive balances borrowed from the physics lab. The weights are
0.951.020.980.971.051.010.981.00 (calculate the mean using these)
ounces. We want to determine a 90% confidence interval for the true mean, µ.
5.Determine z*. (Show your work, including a sketch of the distribution.)
6.Calculate and interpret a 90% confidence interval for the mean weight of the candy bars.
7.Explain the meaning of the confidence level.
8. Determine the sample size you would need to estimate µ within 0.001 at a 90% confidence level.
5. 6. 7. ANSWER: 90% Resulting Confidence Interval for 'true mean': = [0.969, 1.019]
Why???
SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar: Sample mean = 0.994
AVERAGE(0.95, 1.02, 0.98, 0.97, 1.05, 1.01, 0.98)
s: Sample standard deviation = 0.034
n: Number of samples = 7
df: degrees of freedom = 6
Confidence Level = 90
"Look-up" Table ('t-critical value') = 1.943
Look-up Table of ('t critical values') for confidence and prediction intervals. Central two-sided area = 90% with df = 6. Another Look-up method is to utilize Microsoft Excel function: TINV(probability,degrees_freedom) Returns the inverse of the Student's t-distribution 90% Resulting Confidence Interval for 'true mean': x-bar ± ('t critical value') * s/SQRT(n)
= 0.994 ± 1.943 * 0.034/SQRT(7) = [0.969, 1.019]
8. ANSWER: Sample Size = 68 for 90% level of confidence
Why???
SMALL SAMPLE, LEVEL OF CONFIDENCE, NORMAL POPULATION DISTRIBUTION
Margin of Error (half of confidence interval) = 0.001
The margin of error is defined as the "radius" (or half the width) of a confidence interval for a particular statistic.
Level of Confidence = 90
σ: population standard deviation = 0.005
('z critical value') from Look-up Table for 90% = 1.645
The Look-up in the Table for the Standard Normal Distribution utilizes the Table's cummulative 'area' feature. The Table shows positve and negative values of ('z critical') but since the Standard Normal Distribution is symmetric, the magnitude of ('z critical') is important.
For a Level of Confidence = 90% the corresponding LEFT 'area' = 0.45. And due to Table's symmetric nature, the corresponding RIGHT 'area' = 0.45 The ('z critical') value Look-up is 1.645 which means the MIDDLE 'area' = SUM[LEFT 'area' + RIGHT 'area'] for a Level of Confidence = 90.
significant digits = 3
Margin of Error = ('z critical value') * σ/SQRT(n)
n = Sample Size
Algebraic solution for n:
n = [('z critical value') * σ/Margin of Error]²
= [ (1.645 * 0.005)/0.001 ]²
Sample Size = 68 for 90% level of confidence
