# Lower Tail Test of Population Proportion

The null hypothesis of the lower tail test about population proportion can be expressed as follows:

where p0 is a hypothesized lower bound of the true population proportion p.

Let us define the test statistic z in terms of the sample proportion and the sample size:

Then the null hypothesis of the lower tail test is to be rejected if z ≤−zα , where zα is the 100(1 α) percentile of the standard normal distribution.

#### Problem

Suppose 60% of citizens voted in last election. 85 out of 148 people in a telephone survey said that they voted in current election. At 0.5 significance level, can we reject the null hypothesis that the proportion of voters in the population is above 60% this year?

#### Solution

The null hypothesis is that p 0.6. We begin with computing the test statistic.

> pbar = 85/148          # sample proportion
> p0 = .6                # hypothesized value
> n = 148                # sample size
> z = (pbarp0)/sqrt(p0(1p0)/n)
> z                      # test statistic
[1] 0.6376

We then compute the critical value at .05 significance level.

> alpha = .05
> z.alpha = qnorm(1alpha)
z.alpha               # critical value
[1] 1.6449

#### Answer

The test statistic -0.6376 is not less than the critical value of -1.6449. Hence, at .05 significance level, we do not reject the null hypothesis that the proportion of voters in the population is above 60% this year.

#### Alternative Solution 1

Instead of using the critical value, we apply the pnorm function to compute the lower tail p-value of the test statistic. As it turns out to be greater than the .05 significance level, we do not reject the null hypothesis that p 0.6.

> pval = pnorm(z)
> pval                   # lower tail pvalue
[1] 0.26187

#### Alternative Solution 2

We apply the prop.test function to compute the p-value directly. The Yates continuity correction is disabled for pedagogical reasons.

> prop.test(85, 148, p=.6, alt="less", correct=FALSE)

1sample proportions test without continuity
correction

data:  85 out of 148, null probability 0.6
Xsquared = 0.4065, df = 1, pvalue = 0.2619
alternative hypothesis: true p is less than 0.6
95 percent confidence interval:
0.0000 0.63925
sample estimates:
p
0.57432