Question

suppose that in a random selection of 100 colored candies, 22% of them are blue. the candy company claims that the percentage of blue candies is equal to 23%. use a 0.10 significance level to test that claim. identify the null and alternative hypotheses for this test. choose the correct answer below. a. ( h_{0}: p = 0.23 ) ( h_{1}: p>0.23 ) b. ( h_{0}: p = 0.23 ) ( h_{1}: p<0.23 ) c. ( h_{0}: p
eq0.23 ) ( h_{1}: p = 0.23 ) d. ( h_{0}: p = 0.23 ) ( h_{1}: p
eq0.23 ) identify the test statistic for this hypothesis test. the test statistic for this hypothesis test is (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the test statistic in a proportion test

The formula for the test statistic \(z\) in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p\) is the hypothesized proportion, and \(n\) is the sample size.
Given \(\hat{p}=0.22\), \(p = 0.23\), and \(n = 100\).

Step2: Substitute the values into the formula

First, calculate the denominator \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.23\times(1 - 0.23)}{100}}=\sqrt{\frac{0.23\times0.77}{100}}=\sqrt{\frac{0.1771}{100}}=\sqrt{0.001771}\approx0.0421\).
Then, calculate the numerator \(\hat{p}-p=0.22 - 0.23=- 0.01\).
So, \(z=\frac{-0.01}{0.0421}\approx - 0.24\).

Answer:

\(-0.24\)