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Hypothesis test for a proportion

This document is prepared automatically using the following R command.

Problem

Confidence interval of a sample proportion

The approach that we used to solve this problem is valid when the following conditions are met.

Solution

This approach consists of four steps:

1. State the hypotheses

The first step is to state the null hypothesis and an alternative hypothesis.

\[Null\ hypothesis(H_0): P = 0.8\] \[Alternative\ hypothesis(H_1): P \neq 0.8\]

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small..

2. Formulate an analysis plan

For this analysis, the significance level is 0.01`. The test method, shown in the next section, is a one-sample z-test.

2. Select a confidence level.

In this analysis, the confidence level is defined for us in the problem. We are working with a 99% confidence level.

3. Analyze sample data

Using sample data, we calculate the standard deviation (sd) and compute the z-score test statistic (z).

\[sd=\sqrt{\frac{P\times(1-P)}{n}}\] \[sd=\sqrt{\frac{0.8\times(1-0.8)}{100}}=0.04\] \[z=\frac{p-P}{sd}=\frac{0.73-0.8}{0.04}=-1.75\] where \(P\) is the hypothesized value of population proportion in the null hypothesis, \(p\) is the sample proportion, and \(n\) is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic is less than -1.75 or greater than 1.75.

We can use following R code to find the p value.

\[p=pnorm(-abs(-1.75))\times2=0.08\]

Alternatively,we can use the Normal Distribution curve to find p value.

draw_n(z=x$result$z,alternative=x$result$alternative)

4. Interpret results.

Since the P-value (0.08) is greater than the significance level (0.01), we cannot reject the null hypothesis.

Result of propCI()

$data
# A tibble: 1 × 1
  value
  <lgl>
1 NA   

$result
  alpha   n df    p   P   se critical        ME     lower     upper
1  0.01 100 99 0.73 0.8 0.04 2.575829 0.1030332 0.6269668 0.8330332
                      CI     z     pvalue alternative
1 0.73 [99CI 0.63; 0.83] -1.75 0.08011831   two.sided

$call
propCI(n = 100, p = 0.73, P = 0.8, alpha = 0.01)

attr(,"measure")
[1] "prop"

Reference

The contents of this document are modified from StatTrek.com. Berman H.B., “AP Statistics Tutorial”, [online] Available at: https://stattrek.com/hypothesis-test/proportion.aspx?tutorial=AP URL[Accessed Data: 1/23/2022].

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