Supplementary Exercises 6.13 and 6.14 of IPS7e
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6.13:
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Yields of corn in bushels per acre.
Let X_1,...,X_15 denote the 15 yields.

We assume that the observations are i.i.d. (independent and identically
distributed) with mean mu and standard deviation sigma.
We also assume (quite unrealistically) that sigma is known to be 10. 
The average of the X's (Xmean) is 123.80.


(a) With a confidence level of 90%, our zstar-value is the 95% percentile of
N(0,1), which equals 1.645.

90% CI: Xmean +- zstar*sigma/sqrt(n) = 123.8 +- 1.645*10/sqrt(15)
        = 123.8 +- 4.2 = (119.6 , 128)


(b) With a confidence level of 95%, our zstar-value is 1.96:

95% CI: Xmean +- zstar*sigma/sqrt(n) = 123.8 +- 1.96*10/sqrt(15)
        = 123.8 +- 5.1 = (118.7 , 128.9)


(c) With a confidence level of 99%, our zstar-value is the 99.5% percentile
of N(0,1), which equals 2.576:

99% CI: Xmean +- zstar*sigma/sqrt(n) = 123.8 +- 2.576*10/sqrt(15)
        = 123.8 +- 6.7 = (117.1 , 130.5)


(d) The margins of error increase with increasing confidence levels.

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Minitab commands and output for 6.13:

MTB > WOpen "H:\VHM\VHM801\Datasets\Minitab\Chapter 6\ex06_013.mtw".
Retrieving worksheet from file: 'H:\VHM\VHM801\Datasets\Minitab\Chapter
6\ex06_013.mtw'
Worksheet was saved on 26/09/2014

MTB > OneZ 'Yield';
SUBC>   Sigma 10;
SUBC>   Confidence 90;
SUBC>   Alternative 0.
One-Sample Z: Yield 

The assumed standard deviation = 10

Variable   N    Mean  StDev  SE Mean       90% CI
Yield     15  123.80  12.26     2.58  (119.55, 128.05)

MTB > OneZ 'Yield';
SUBC>   Sigma 10;
SUBC>   Confidence 95;
SUBC>   Alternative 0.
One-Sample Z: Yield 

The assumed standard deviation = 10

Variable   N    Mean  StDev  SE Mean       95% CI
Yield     15  123.80  12.26     2.58  (118.74, 128.86)

MTB > OneZ 'Yield';
SUBC>   Sigma 10;
SUBC>   Confidence 99;
SUBC>   Alternative 0.
One-Sample Z: Yield 

The assumed standard deviation = 10

Variable   N    Mean  StDev  SE Mean       99% CI
Yield     15  123.80  12.26     2.58  (117.15, 130.45)


6.14:
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We assume (hypothetically) that the Xmean=123.80 came from a sample 
of size 50.


(a) With a confidence level of 95%, our zstar-value is 1.96, and it's
the same formula as in (b) of the previous exercise:

95% CI: Xmean +- zstar*sigma/sqrt(n) = 123.8 +- 1.96*10/sqrt(50)
        = 123.8 +- 2.8 = (121.0 , 126.6)


(b) The margin of error for n=50 (2.8) is narrower than the margin of
error for n=15 (5.1). This is because a larger sample size provides 
more precise information about the unknown parameter (here, the population 
mean). Generally speaking, margin of errors always decrease with
increasing sample size.


(c) The margin of errors for 90% and 99% intervals will also be narrower
for n=50 than for n=15, for the reasons just explained above. The ratio
between the margin of errors obtained at sample sizes 15 and 50 will be 
sqrt(50)/sqrt(15)=1.825, so that the interval for n=15 is 1.825 times
wider than that for n=50, regardless of the confidence level.

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Minitab commands and output for 6.14:

MTB > OneZ 50 123.8;
SUBC>   Sigma 10;
SUBC>   Confidence 95;
SUBC>   Alternative 0.
One-Sample Z 

The assumed standard deviation = 10

 N    Mean  SE Mean       95% CI
50  123.80     1.41  (121.03, 126.57)