Download Homework #4 STAT651

Homework #4 STAT651
1) Open up the DNA adduct level data. Note and write down the variable names.
- diet, proximal, and distal
2) Construct summary statistics for each of the two diets.
Descriptives
DNA adduct level
in dis tal colon
Diet Group
Corn Oil Diet
Fish Oil Diet
Mean
95% Confidence
Interval for Mean
5% Trimmed Mean
Median
Variance
Std. Deviation
Minimum
Maximum
Range
Interquartile Range
Skewness
Kurtos is
Mean
95% Confidence
Interval for Mean
5% Trimmed Mean
Median
Variance
Std. Deviation
Minimum
Maximum
Range
Interquartile Range
Skewness
Kurtos is
Lower Bound
Upper Bound
Lower Bound
Upper Bound
Statis tic
29.5971
21.1720
Std. Error
3.92816
38.0222
28.6555
26.3683
231.457
15.21371
11.63
64.52
52.89
22.2349
.956
.638
16.7633
12.4187
.580
1.121
2.02564
21.1079
16.2711
16.2445
61.548
7.84528
4.82
37.57
32.75
9.3444
1.068
2.717
.580
1.121
a) Do you see any indication that there might be a massive outlier?
- Not at all. The corn oil group had higher means, medians,
standard deviations and interquartile ranges than the fish oil
group. This general consistency is something that one might not
see if there were a massive outlier.
3) Now construct the boxplot for the two populations.
70
60
50
40
28
30
20
10
0
N =
15
15
Corn Oil Diet
Fish Oil Diet
Diet Group
a) Compare the two populations in terms of their measures of central tendency,
i.e., their sample median.
- The sample median from ‘Corn Oil Diet’ group is 26.36 and one from ‘Fish Oil
Diet’ group 16.24. It looks that the fish oil enhanced diet group had less averagedamage(smaller adduct levels) than the corn oil enhanced diet group.
b) Compare the two populations in terms of their measures of variability, i.e. the
interquartile ranges(IQR).
- The IQR from the ‘Corn Oil Diet’ group is 22.23 and one from ‘Fish Oil Diet’
group 9.33. The ‘Corn Oil Diet’ group has larger variability than ‘Fish Oil Diet’
group does.
4) (Fish Oil diet) the population standard deviation  = 8. Use the empirical rule to
narrow down with 95% confidence where the population mean for this diet group
is.
- The general rule is that 95% of the population values are within 1.96
standard deviations = 1.96 * 8 = 15.86 of the sample mean (16.76). Hence,
95% of the population values should lie between 16.76 – 15.86 = 0.90 and
16.76 + 15.86 = 32.62.
- On the other hand, the chance is 95% that the population mean is within
1.96 standard errors of the sample mean.
- Since the population standard deviation is known, the standard error is
1.96 *  / (square root of n) = 1.96 * 8 / (square root of 15) = 15.86 / 3.87
= 4.10.
- Hence, with 95% probability, the population mean is between 16.76 – 4.10
= 12.66 and 16.76 + 4.10 = 20.86.
- By the way, the SPSS output notes that the exact confidence interval for
the population mean is 12.41 and 21.10, once one takes into account the
fact that the population standard deviation is unknown. This interval is
slightly longer because we have to estimate the population standard
deviation by the sample standard deviation.
5) The population mean  = 17. What percentage of the population of rats who are
fed a fish-oil enhanced diet will have a DNA adduct level exceeding 33?
- Pr(X > 33) = Pr[ (X – 17)/8 > (33 – 17)/8 ] = Pr[Z > 2 ] = 0.0228.
6) Let X stand for the number of red petals of bluebonnets that are near the highway.
Suppose that these bluebonnets have a population mean  = 2.8 and a population
standard deviation = 2. Compute the following probabilities. Round the z-score
to 2 digits. If the z-score is greater than 3.4, set it = 3.4. If the z-score is less than 3.4, set it equal to -3.4.
- Pr(X > 4.8)
(Solution) Pr( X > 4.8 ) = Pr[ (X - ) /  > (4.8 - ) /  ]
= Pr[ Z > ( 4.8 – 2.8 ) / 2 ]
= Pr[ Z > 1.0 ]
= 1 – Pr[Z  1.0] = 1 - 0.8413 = 0.1587
- Pr(X < 4.8)
(Solution) Pr( X < 4.8 ) = Pr[ (X - ) /  < (4.8 - ) /  ]
= Pr[ Z < ( 4.8 – 2.8 ) / 2 ]
= Pr[ Z < 1.0 ]
= 0.8413
- Pr(X < 2.8)
(Solution) Pr( X < 2.8 ) = Pr[ (X - ) /  < (2.8 - ) /  ]
= Pr[ Z < ( 2.8 – 2.8 ) / 2 ]
= Pr[ Z < 0.0 ]
= 0.5
7) Let X stand for the number of red petals of bluebonnets that are far from the
highway. Suppose that these bluebonnets have a population mean  = 4.8 and a
population standard deviation = 2. Compute the following probabilities. Round
the z-score to 2 digits. If the z-score is greater than 3.4, set it = 3.4. If the z-score
is less than -3.4, set it equal to -3.4.
- Pr(X > 4.8)
(Solution) Pr( X > 4.8 ) = Pr[ (X - ) /  > (4.8 - ) /  ]
= Pr[ Z > ( 4.8 – 4.8 ) / 2 ]
= Pr[ Z > 0.0 ]
= 0.5
- Pr(X < 4.8)
(Solution) Pr( X < 4.8 ) = Pr[ (X - ) /  < (4.8 - ) /  ]
= Pr[ Z < ( 4.8 – 4.8 ) / 2 ]
= Pr[ Z < 0.0 ]
= 0.5
- Pr(X < 2.8)
(Solution) Pr( X < 2.8 ) = Pr[ (X - ) /  < (2.8 - ) /  ]
= Pr[ Z < ( 2.8 – 4.8 ) / 2 ]
= Pr[ Z < - 1.0 ]
= 0.1587