Solutions Homework 1 for Math 3200 PDF

Preview

Summary

Solutions Homework 1 for Math 3200. Homework Problems...

Description

Problem 1 (Section 1.4 (#3)) Solution: a) Univariate b) Quantitative c) Statistical Population is {x1 , · · · , xn } where xi is the total number of engine and transmission nonconformances of the ith car. d) Bivariate

Problem 2 (Section 1.5 (#4)) Solution: • Figure 1 gives the scatterplot matrix of the data. As is evident from the plots, latitude seems to be a better predictor of the temperatures, since there seems to be a clear decreasing pattern in the Temperature vs Latitude plot, while no visible pattern exists with change in longitude.

Figure 1: Scatterplot of the Data • The 3-D scatterplot of the data is given in Figure 2. This plot also indicates that latitude is a better predictor for the temperature, since there seems to be clear decreasing correlation between January Temperature and Latitude.

1

Figure 2: Caption

Problem 3 (Section 1.6 (#9)) Solution: a) 21 1 {1 + 2 + 3 + 4 + 5 + 6} = = 3.5 6 6 = 2.92 (Check!)

µX = 2 σX

b) We replicated the process 1000 times and calculated mean and variance for each sample. Suppose the means and variances of the samples be mX and vx2 respectively. The histogram of mX −3.5 and vX −2.92 are given in Figure 3. From the figure, it seems that the sample mean and sample variance approximate the population mean and population variance reasonably well. c) For one sample, the sample proportion is (0.15, 0.17, 0.16, 0.23, 0.18, 0.11). However, to check whether they are reasonable close to 16 , we have to take many such samples, like we did in last part. The histogram of each of the proportions subtracted by 1/6 is given in Figure 4. From the histograms, it seems that the sample proportions estimates are reasonable close to 16 .

2

Figure 3: Histogram of Sample Mean Error and Sample Variance Error

Figure 4: Histogram of Sample Proportion Error 3

Problem 4 (Section 1.6 (#13)) Solution: a) For mean: y¯ =

Pn

i=1 yi

n P n nc1 + i=1 xi = n = c1 + x ¯

For Variance: Sy2

Pn

− y¯)2 n−1 Pn 2 {c1 + xi − (c1 + x ¯)} = i=1 n−1 = S x2

=

i=1 (yi

From this, Sy = Sx , by taking square root. b) For mean: Pn

i=1 yi

y¯ =

Pnn

i=1 c2 xi

=

n

= c2 x ¯ For Variance: Sy2

Pn

− y¯)2 n−1 Pn 2 {c2 xi − c2 x ¯} = i=1 n−1 Pn 2 2 {x − x ¯} c = i=1 2 i n−1 2 2 = c2 Sx

=

i=1 (yi

From this, Sy = |c2 |Sx , by taking square root. c) For mean: y¯ =

Pn

i=1 yi

Pnn

+ c 2 xi ) n = c1 + c2 x ¯ =

For Variance:

i=1 (c1

Pn

− y¯)2 n−1 Pn 2 {c1 + c2 xi − (c1 + c2 x ¯)} = i=1 n−1 Pn 2 2 ¯} c {xi − x = i=1 2 n−1 = c22 Sx2

Sy2 =

i=1 (yi

From this, Sy = |c2 |Sx , by taking square root.

4

Problem 5 (Section 1.7 (#3)) Solution: a) The summary of the dataset is given in Figure 5.

Figure 5: Summary of the Dataset b) The 90th percentile is 31.068. (Code: quantile(t2, 0.9)) c) The boxplot is given in Figure 6.(Code: boxplot(t2)). There are no outliers.

Figure 6: Boxplot

Problem 6 (Section 1.8 (#11)) Solution: a) Comparative Bar Graph is given in Figure 7. b) Stacked Bar Graph is given in Figure 8. c) The comparative bar graph is better for comparing the volume of sales for online and catalog, while stacked bar graph is better for showing variation in the total volume and sales.

5

Figure 7: Comparative Bar Graph

Figure 8: Stacked Bar Graph

m=rbind(oc$Online, oc$Catalog) barplot(m, names.arg=oc$Month, ylim=c(0, 70), col=c("darkblue", "red"), legend.text= c("Online", "Catalog"), beside=TRUE, las=2) # constructs the bar graph barplot(m, names.arg= oc$Month, ylim=c(0, 100), col=c("red", "blue"), las =2) legend("topright", pch=c(22, 22), col=c("red", "blue"), legend=c("Online","Catalog")) 6...