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Solutions to the problems in Test #3
Math 146
12/4/15
Data Set #1
0.64531 1.94499 1.32485 1.97105
0.86830 0.59840 0.68198 0.91983
1.54369 0.82731 0.36510 0.37426
0.66417 1.26636 0.77177 1.37512
0.48117 1.02087 0.93507 0.71663
Sum
19.2962
Count
20
Sum of squares
22.7967
Data Set #2
Mean
0.79618
Sample Variance
0.02053
Sum
23.8855
Count
30
Descriptive Statistics: Calculate for data set #1 only
•
Sample mean, sample standard deviation, and sample variance
•
1st Quartile, Median, and 3rd Quartile
•
Minimum, Maximum, and Range
Solutions
Mean
0.96481
Median
0.84781
Standard Deviation 0.46901
Sample Variance
0.21997
Range
1.60596
Minimum
0.36510
Maximum
1.97105
1 quartile
0.65945
3 quartile
1.28098
Inferential Statistics
Confidence Interval for data set #1
Choosing a confidence level, use the previous data for a confidence interval for the “true” mean (the
variance is unknown).
If you have lots of time left, you may also construct a confidence interval for the second data set.
data 1
90% CI for the Mean from 0.78347
to
1.14615
data 1
95% CI for the Mean from 0.74531
to
1.18431
data 1
99% CI for the Mean from 0.66477
to
1.26485
data 2
90% CI for the Mean from 0.75173
to
0.84064
data 2
95% CI for the Mean from 0.74268
to
0.84969
data 2
99% CI for the Mean from 0.67497
to
0.86175
Testing for the mean for data set #2
Test the hypothesis
H0: μ = 0.75
H0: μ ≠ 0.75
t Stat
1.76534
p Value 0.04402
The p-value is about 4.4%, which means that you would reject the Null Hypothesis at the 10% (this is a
two-tailed test) significance level, but not at any level below 8.8%.
Testing for equality of two means (clearly, this refers to both data sets)
We now want to check if the two data sets could come from “populations” (i.e., probabilistic models)
with the same “true” mean.
Set up a test, without assuming that the variances are equal.
Hypothesized Mean Difference
0
Observed Mean Difference
0.16863
df
21.3840
t Stat
1.56010
p-value
0.06671
t Critical two-tail
2.07734
The “t Critical two-tailed” refers to a 5% significance, and we see that we would not reject at this level.
In fact, this being a 2-tailed test, a p-value of 6.7% means that the lowest significance at which we
would reject the Null Hypothesis would be twice this value, or 13.4% – an unusually high choice.
If you are curious, the computer used the following means and standard deviations for its simulations:
Data Set μ
σ
Set #1
1
0.4
Set #2
0.8
0.2