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Unformatted text preview: ce interval estimate of the mean 12month weight gain for all
quitters.
Using margin of error:
Population: Population:
std dev σ: 6 Interval Estimation Sample: std dev σ: 6 Sample:
mean: 13.15384615
sample size n:
13 mean: =AVERAGE(A2:A16)
sample size n: =COUNT(A2:A16) Sampling Distribution:
std error: 1.664100589 Sampling Distribution: see Smoking.xlsx Example 2
One of the few negative side effects of quitting smoking is weight gain.
Suppose that the weight gain in the 12 months following a cessation in
smoking is normally distributed with a standard deviation of 6 kilograms.
To estimate the mean weight gain, a random sample of 13 quitters was
drawn their weight gains recorded and listed here. Determine the 90%
confidence interval estimate of the mean 12month weight gain for all
quitters.
Using CONFIDENCE function:
Population: Population: Interval Estimation std dev σ: 6 6 mean: std dev σ: =AVERAGE(A2:A14) Sample:
Sample:
mean: 13.15384615
Interval Estimate: see Smoking.xlsx lower:
Interval Estimate: =E6CONFIDENCE.NORM(0.1,E3,COUNT(A2:A14)) upper: =E6+CONFIDENCE.NORM(0.1,E3,COUNT(A2:A14)) To Try…
1. 2. Suppose that shopping times for customers at a local grocery store
are normally distributed. A random sample of 16 shoppers in the local
grocery store had a mean time of 25 minutes. Assume σ=6 minutes.
Find the 95% confidence interval for the population mean. Interpret
your results.
A process produces bags of refined sugar. The weights of the
contents of these bags are normally distributed with standard
deviation 1.2 ounces. The contents of a random sample of 25 bags
had a mean weight of 19.8 ounces. Find the upper and lower
confidence limits of a 99% confidence interval for the true mean
weight for all bags of sugar produced by the process. If the bags are
intended to have 20 ounces of sugar, comment on the filling process. Interval Estimation To Try…
Suppose that shopping times for customers at a local grocery store
are normally distributed. A random sample of 16 shoppers in the local
grocery store had a mean time of 25 minutes. Assume σ=6 minutes.
Find the 95% confidence interval for the population mean. Interpret
σ
your results.
Confidence interval:
1. x ±z 2
α n Population:
std dev σ: 6 Interval Estimation confidence level = 95%, therefore, α=.05
zα/2 = z.05/2 = 1.960
confidence interval:
Sample:
σ
x ± 1.960
size n:
16
n
mean xbar:
25
6
25 ± 1.960
16
25 ± 2.94
Interval Estimate:
22.06 ≤ µ ≤ 27.94
significance α:
0.05
Or, using MS Excel, CONFIDENCE.NORM(0.05,6,16)=2.939945977
confidence:
95%
We are 95% confident that the mean shopping time for customers at the grocery
store is between 22 and 28 minutes.
Calulations: To Try…
A process produces bags of refined sugar. The weights of the
contents of these bags are normally distributed with standard
deviation 1.2 ounces. The contents of a random sample of 25 bags
had a mean weight of 19.8 ounces. Find the upper and lower
confidence limits of a 99% confidence interval for the true mean
weight for all bags of sugar produced by the process. If the bags are
intended to have 20 ounces of sugar, comment on the filling
σ
Population:
process.
Confidence interval: x ±z
2. α2 n Interval Estimation confidence level = 99%, therefore, α=.01
zα/2 = z.01/2 = 2.576
1.2
confidence interval: 19.8 ± 2.576
25
19.8 ± 0.61824
19.18176 ≤ µ ≤ 20.41824 std dev σ: 1.2 size n: 25 mean xbar: 19.8 Sample: Interval Estimate: Or, using MS Excel,
CONFIDENCE.NORM (0.01,1.2,25)=0.618199033
significance α:
0.01
We are 99% confident that the mean weight of the bags of sugar is between
confidence:
99%
19.2 and 20.4 ounces. Because 20 ounces is in the interval, we can assume the
filling process is working correctly....
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This document was uploaded on 03/05/2014.
 Spring '13

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