Section 8.1: Inference for a Single Proportion

Example: Changing Majors

Below are the commands and answers for the in-class handout.

Question 1

(a)

Find the proportion of students in the sample who changed majors:

399/950
## [1] 0.42

So \(\hat{p}=.042\).

(b)

Find the standard error for \(\hat{p}\):

sqrt(.42*(1-.42)/950)
## [1] 0.01601315

So \(SE_{\hat{p}}=.016\).

(c)

Find \(z^*\) for the 90% confidence interval:

xqnorm(.95)
##  P(X <= 1.64485362695147) = 0.95
##  P(X >  1.64485362695147) = 0.05

## [1] 1.644854

So \(z^*=1.645\).

(d)

Find the margin of error for the 90% confidence interval:

1.645*.016
## [1] 0.02632

So \(m=.02632\)

(e)

Find the 90% confidence interval for \(p\):

.42-.02632
## [1] 0.39368
.42+.02632
## [1] 0.44632

The 90% confidence interval for \(p\) is \((.39368, .44632)\).

(f)

The 90% confidence interval in terms of percentages is \((39.368%, 44.632%)\).

(g)

.39368*37500
## [1] 14763
.44632*37500
## [1] 16737

The 90% confidence interval in terms of number of students is \((14763, 16737)\).

(h)

We will use the formula \(\left(\frac{z^*}{m}\right)^2p^*(1-p^*)\), where \(p^*\) is a guessed value for the proportion of successes in the future sample. For this example we can use \(p^*=\hat{p}=.42\):

(1.645/.02)^2*.42*(1-.42)
## [1] 1647.969

We should use a sample of \(1648\) students to obtain a margin of error of approximately \(.02\).

Question 2

(a)

\(H_0: p=.45\)

\(H_a: p<.45\)

(b)

Compute the \(z\) statistic:

(.42-.45)/sqrt(.45*(1-.45)/950)
## [1] -1.858641

So \(z=-1.85641\).

(c)

Find the corresponding \(P\)-value:

xpnorm(-1.859)
## 
## If X ~ N(0,1), then 
## 
##  P(X <= -1.859) = P(Z <= -1.859) = 0.0315
##  P(X >  -1.859) = P(Z >  -1.859) = 0.9685

## [1] 0.03151357

The \(P\)-value is \(.0315\).

(d)

The \(P\)-value is less than \(.05\), which means that we have enough evidence at the 5% significance level supporting the claim of the tour guide.

Question 3

(a)

prop.test(399, 950, p=.45, alternative="less")
## 
##  1-sample proportions test with continuity correction
## 
## data:  399 out of 950
## X-squared = 3.3344, df = 1, p-value = 0.03392
## alternative hypothesis: true p is less than 0.45
## 95 percent confidence interval:
##  0.0000000 0.4470594
## sample estimates:
##    p 
## 0.42

The \(P\)-value from the test is different from the \(P\)-value we found.

prop.test(399, 950, p=.45, conf.level=.9)
## 
##  1-sample proportions test with continuity correction
## 
## data:  399 out of 950
## X-squared = 3.3344, df = 1, p-value = 0.06785
## alternative hypothesis: true p is not equal to 0.45
## 90 percent confidence interval:
##  0.3934041 0.4470594
## sample estimates:
##    p 
## 0.42

The 90% confidence interval from the test is also not quite the same as the interval we found.

(b)

Repeat the test with the continuity correction turned off:

prop.test(399, 950, p=.45, alternative="less", correct=F)
## 
##  1-sample proportions test without continuity correction
## 
## data:  399 out of 950
## X-squared = 3.4545, df = 1, p-value = 0.03154
## alternative hypothesis: true p is less than 0.45
## 95 percent confidence interval:
##  0.00000 0.44653
## sample estimates:
##    p 
## 0.42

The \(P\)-value in the output is the same as the value we found in question 2.

prop.test(399, 950, p=.45, correct=F)
## 
##  1-sample proportions test without continuity correction
## 
## data:  399 out of 950
## X-squared = 3.4545, df = 1, p-value = 0.06308
## alternative hypothesis: true p is not equal to 0.45
## 95 percent confidence interval:
##  0.3889986 0.4516458
## sample estimates:
##    p 
## 0.42

The 90% confidence interval in the output is the same as the one we found in question 1.

(d)

We can use the \(P\)-value from the test inside the xqnorm command to find the value of the \(z\) statistic.

xqnorm(.03154)
##  P(X <= -1.85862719226929) = 0.03154
##  P(X >  -1.85862719226929) = 0.96846

## [1] -1.858627

So \(z=-1.8586\) which is almost the same as in question 2.