Sangria: You need a 95% confidence interval around the mean of the amount of Tri

Question

Sangria: You need a 95% confidence interval around the mean of the amount of Triple-Sec you put in your Sangria. You have sampled 50 batches and your sample mean is 8.2 02. You know from precisely measuring every batch you've ever made that the standard deviation is 0.5. Find the following: Beer: You are the Quality Control engineer for a micro-brewery startup. You do not have any historical data. You need a 90% confidence interval around the mean concentration of alcohol in your beer. You sample 35 beers and calculate a sample mean of 5.6% with a sample standard deviation of 1.2 t or z value What would happen to this interval if you sampled 50 beers?

Explanation / Answer

Q1.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=8.2
Standard deviation( sd )=0.5
Sample Size(n)=50
Margin of Error = Z a/2 * 0.5/ Sqrt ( 50)
= 1.96 * (0.071)
= 0.139

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=8.2
Standard deviation( sd )=0.5
Sample Size(n)=50
Confidence Interval = [ 8.2 ± Z a/2 ( 0.5/ Sqrt ( 50) ) ]
= [ 8.2 - 1.96 * (0.071) , 8.2 + 1.96 * (0.071) ]
= [ 8.061,8.339 ]

[ANSWERS]
Z, 1.96, ME = 0.139, LL = 8.061, UL = 8.339

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