Question

1. Consider the number of days absent from a random sample of six students during a semester: A= {2, 3, 2, 4, 2, 5}
   1. Compute the arithmetic mean, geometric mean, median, and mode by hand and verify the results using R
      1. Arithmetic Mean: X=i=1nXin=2+3+2+4+2+56=3

mean(data2$absent)

[1] 3

2. Geometic Mean: GMx=Πi=1nX11n=2∙3∙2∙4∙2∙516=2.79816

>gmean <- prod(data2$absent)^(1/length(data2$absent))

> gmean

[1] 2.798166

3. Median: X=12n+1th, Xi2,2,2,3,4,5, n=6=126+1th ranked value=3.5, value=2.5 days absent

>median(data2$absent)

[1] 2.5

4. Mode: Most frequent value=2

> mode <- names(table(data2$absent)) [table(data2$absent)==max(table(data2$absent))]

> mode

[1] "2"

2. Compute the variance, standard deviation, and coefficient of variation by hand and verify the results using R.
   1. Sample Variance: S2=i=1nXi-X2n-1=2-32+3-32+2-324-322-325-326-1=85=1.6

> var(data2$absent)

[1] 1.6

2. Sample Standard Deviation: s=s2=i=1nXi-X2n-1=1.6=1.26491

> sd(data2$absent)

[1] 1.264911

3. Coefficient of Variation: CV=SX=1.264912.5=.421637

> cv <- abs(sd(data2$absent)/mean(data2$absent))

> cv

[1] 0.421637

3. Compute the interquartile range of this data by hand and with R

> IQR(data2$absent, type=6)

[1] 2.25

4. Display the data above with a well‐formatted and well‐labeled pie chart created with R.  Recall that with a pie chart, the slices sum to 100 percent, so the slice labeled “2 days absent” should be 50 percent of the pie.

Pie Chart with Percentages

slices <- c(50, 16.6666667, 16.6666667, 16.6666667)

lbls <- c("2 days absent", "3 days absent", "4 days absent", "5 days absent")

pct <- round(slices/sum(slices)*100)

lbls <- paste(lbls, pct) # add percents to labels

lbls <- paste(lbls,"%",sep="") # ad % to labels

pie(slices,labels = lbls, col=rainbow(length(lbls)),

    main="Pie Chart of Absences")

5. Display the data above with a well‐labeled and well‐formatted frequency distribution (histogram) using R.

x <- data2$absent

> h<-hist(x, breaks=10, col="magenta", xlab="# of student absences",

+         main="Histogram of student absences")

> xfit<-seq(min(x),max(x),length=40)

> yfit<-dnorm(xfit,mean=mean(x),sd=sd(x))

> yfit <- yfit*diff(h$mids[1:2])*length(x)

> lines(xfit, yfit, col="blue", lwd=2)

6. Use the relationship between the mean and median to determine whether the distribution above is symmetrical, negatively skewed or positively skewed.  Explain.
   1. Mean > Median = 3 > 2.5, the distribution above is positively skewed.
7. Calculate excess kurtosis using the formula from lecture.  Is this distribution leptokurtic, mesokurtic, or platykurtic?  Explain.  Compared to the normal distribution with the same variance, does this distribution exhibit fatter tails?  Briefly explain
   1. 1ni=1nxi-X41ni=1nxi-X22=162-34+3-34+2-34+4-34+2-34+5-34162-32+3-32+2-32+4-32+2-32+5-322=103169=158≈1.875
   2. Excess kurtosis = kurtosis – 3 = 1.875 – 3 = -1.125, -1.125 < 3
   3. This distribution is platykurtic as excess kurtosis < 0, kurtosis < 3
   4. No it will exhibit thinner tails than a normal distribution.

Suppose someone claims that the population mean is 5 days absent (?: ? =5). What is the alternative hypothesis?  Can you reject the null hypothesis at the 5‐percent and 1‐percent levels of significance?  Use the critical value, p‐value, and confidence interval approaches both “by hand” and with R.  In the “by hand” approach, you can use R (or a statistical table) to get the critical values. ????

Answer 1

A= {2, 3, 2, 4, 2, 5}

Null Hypothesis H0: ? =5

Alternative Hypothesis H1; ? 5

Sample mean = (2 + 3 + 2 + 4 + 2 + 5) / 6 = 3

Sample variance = [ (2 - 3)² +  (3 - 3)² +  (2 - 3)² +  (4 - 3)² +  (2 - 3)² +  (5 - 3)² ] / 5

= 1.6

Sample Standard deviation, s = = 1.264911

Standard error, SE = s / = 1.264911 / = 0.5163978

Test statistic, t = (3 - 5) / 0.5163978 = -3.872983

Degree of freedom = n-1 = 6 - 1 = 5

For two tail test, P-value = 2 * P(t < -3.872983) = 0.01172482

P-value approach - Since, p-value is less than 0.05 significance level, we reject null hypothesis H0 at 5% significance level and conclude that there is strong evidence that ? 5.

Since, p-value is greater than 0.01 significance level, we fail to reject null hypothesis H0 at 1% significance level and conclude that there is no strong evidence that ? 5.

Critical value of t at df = 5 and 5‐percent and 1‐percent levels of significance are  2.57 and 4.032. That is we reject H0 if t <-2.57 or t > 2.57 at 5% significance level. And we reject H0 if t <-4.032 or t > 4.032 at 1% significance level.

Critical value approach -

Since, test statistic lie in the critical region we reject null hypothesis H0 at 5% significance level and conclude that there is strong evidence that ? 5.

Since, test statistic does lie in the critical region, we fail to reject null hypothesis H0 at 1% significance level and conclude that there is no strong evidence that ? 5.

For 5% significance level ,

Margin of error = t * SE = 2.57 * 0.5163978 = 1.327142

95% confidence interval is,

(3 - 1.327142, 3 + 1.327142)

(1.672858, 4.327142)

Since hypothesized mean 5 does not lie in the 95% confidence interval, we conclude that there is strong evidence that ? 5.

For 1% significance level ,

Margin of error = t * SE = 4.032 * 0.5163978 = 2.082116

99% confidence interval is,

(3 - 2.082116, 3 + 2.082116)

(0.917884, 5.082116)

Since hypothesized mean 5 lie in the 99% confidence interval, we conclude that there is no strong evidence that ? 5.

Using R,

For 5% significance level ,

t.test(A, mu = 5)

   One Sample t-test

data: A
t = -3.873, df = 5, p-value = 0.01172
alternative hypothesis: true mean is not equal to 5
95 percent confidence interval:
1.672557 4.327443
sample estimates:
mean of x
3

For 1% significance level ,

t.test(A, mu = 5, conf.level = 0.99)

   One Sample t-test

data: A
t = -3.873, df = 5, p-value = 0.01172
alternative hypothesis: true mean is not equal to 5
99 percent confidence interval:
0.9178103 5.0821897
sample estimates:
mean of x
3