Quiz 6 Scripts

1. Study the lecture notes and screen cast for Statistical Tests and Confidence Intervals, Part 1: One Mean or the Difference of Two Means (10:16). Consider testing H0:μ=7 against H1:μ>7, given this random sample from a normal population: 7, 12.2, 7.1, 10.1, 11, 14.7, 8.8, 6, 11.4, 13.8, 13.4, 9.2, 17.4, 13.1, 6.5, 9.8, 5.4, 11.4, 11, 10.8, 1.5, 22. Find the value of the test statistic, t.

x = c(7, 12.2, 7.1, 10.1, 11, 14.7, 8.8, 6, 11.4, 13.8, 13.4, 9.2, 17.4, 13.1, 6.5, 9.8, 5.4, 11.4, 11, 10.8, 1.5, 22)
t.test(x, alternative="greater", mu =7)
# The t value is 3.8853

Consider testing H0:μ=5 against H1:μ>5, given this random sample from a normal population: 4.3, 2.6, 12.2, 7.3, 6.8, 4.8, 1, 5.2, 12.5, 5.6, 3.5, 11.9, 6.4, 4.4, 6.3, 11.3, 7.3, 7.4, 0.9, 11, 9. Find the value of the test statistic, t.

x = c(4.3, 2.6, 12.2, 7.3, 6.8, 4.8, 1, 5.2, 12.5, 5.6, 3.5, 11.9, 6.4, 4.4, 6.3, 11.3, 7.3, 7.4, 0.9, 11, 9)
t.test(x, alternative = "greater", mu=5)
# t = 2.2716

2. Assuming H0 is true, find the probability of seeing a test statistic as extreme as the one you found. (That is, find the P-value.)

x = c(7, 12.2, 7.1, 10.1, 11, 14.7, 8.8, 6, 11.4, 13.8, 13.4, 9.2, 17.4, 13.1, 6.5, 9.8, 5.4, 11.4, 11, 10.8, 1.5, 22)
t.test(x, alternative="greater", mu =7)
# The P value is 0.0004272

2. Assuming H0 is true, find the probability of seeing a test statistic as extreme as the one you found. (That is, find the P-value.)

x = c(4.3, 2.6, 12.2, 7.3, 6.8, 4.8, 1, 5.2, 12.5, 5.6, 3.5, 11.9, 6.4, 4.4, 6.3, 11.3, 7.3, 7.4, 0.9, 11, 9)
t.test(x, alternative = "greater", mu=5)
# p = 0.01715

3. Find left endpoint of a 90% (two-sided) confidence interval for the unknown population mean, μ, given the previous random sample.

x = c(4.3, 2.6, 12.2, 7.3, 6.8, 4.8, 1, 5.2, 12.5, 5.6, 3.5, 11.9, 6.4, 4.4, 6.3, 11.3, 7.3, 7.4, 0.9, 11, 9)
t.test(x, alternative = "two.sided", conf.level = .90)
# left is 5.420721

4. Find the right endpoint of the previous confidence interval.

x = c(4.3, 2.6, 12.2, 7.3, 6.8, 4.8, 1, 5.2, 12.5, 5.6, 3.5, 11.9, 6.4, 4.4, 6.3, 11.3, 7.3, 7.4, 0.9, 11, 9)
t.test(x, alternative = "two.sided", conf.level = .90)
# right is 8.074517

5. Study Part 2: F Test for Equality of Variances (3:52). Consider testing H0:σ2X=σ2Y against H1:σ2X≠σ2Y from two independent samples from normal populations with unknown means μX and μY and standard deviations σX and σY. The X’s are 7.9, 19.4, 15, 13.5, 6.5, 5.1. The Y’s are 8.4, 5.3, 1.5, 3.8, 6.1, 7.4, 3.5. Find the value of the test statistic.

X = c(7.9, 19.4, 15, 13.5, 6.5, 5.1)
Y = c(8.4, 5.3, 1.5, 3.8, 6.1, 7.4, 3.5)
ftestout = var.test(X, Y, ratio = 1); ftestout
# F = 5.4794

6. Find the P-value.

X = c(7.9, 19.4, 15, 13.5, 6.5, 5.1)
Y = c(8.4, 5.3, 1.5, 3.8, 6.1, 7.4, 3.5)
ftestout = var.test(X, Y, ratio = 1); ftestout
# p-value = 0.06129

7. Study Part 3: Chi-Squared Tests (7:45). Consider testing whether the row and column variables are independent, given the following two-way table of observed counts: 11 16 13 10 14 19 19 13 15 Find the value of the chi-square test statistic.

m = matrix(data = c(11, 16, 13, 10, 14, 19, 19, 13, 15), nrow = 3, ncol = 3, byrow = TRUE);m

     [,1] [,2] [,3]
[1,]   11   16   13
[2,]   10   14   19
[3,]   19   13   15

c = chisq.test(m); c$statistic

X-squared 
 4.500904 

8. Find the P-value.

m = matrix(data = c(11, 16, 13, 10, 14, 19, 19, 13, 15), nrow = 3, ncol = 3, byrow = TRUE);m

     [,1] [,2] [,3]
[1,]   11   16   13
[2,]   10   14   19
[3,]   19   13   15

c = chisq.test(m); c$p.value

[1] 0.3424403

9. Study Part 4: One Proportion or the Difference of Two Proportions (8:46). Consider testing H0:p=0.5 against H1:p≠0.5, given that there were 35 successes in a sample of size 52. Find the value of the (chi-squared) test statistic. (Do not use the continuity correction.)

x = 35
n = 52
p = 0.5
prtest = prop.test(x, n, p, correct = FALSE); prtest$statistic

X-squared 
 6.230769 

prtest

    1-sample proportions test without continuity correction

data:  x out of n, null probability p
X-squared = 6.2308, df = 1, p-value = 0.01255
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.5375619 0.7847793
sample estimates:
        p 
0.6730769 

# X-squared = 5.557692

10. Find the P-value.

x = 35
n = 52
p = 0.5
prtest = prop.test(x, n, p); prtest$p.value

[1] 0.01839965