Question

suppose that in a random selection of 100 colored candies, 24% of them are blue. the candy company claims that the percentage of blue candies is equal to 27%. use a 0.05 significance level to test that claim.

d. ( h_0: p = 0.27 )
( h_1: p>0.27 )
identify the test statistic for this hypothesis test.
the test statistic for this hypothesis test is - 0.68
(round to two decimal places as needed.)
identify the p - value for this hypothesis test.
the p - value for this hypothesis test is
(round to three decimal places as needed.)

Explanation:

Step1: Determine the type of test

This is a two - tailed test because the claim is \(p = 0.27\) (the null hypothesis \(H_0:p = 0.27\) and the alternative hypothesis \(H_1:p
eq0.27\)). The test statistic \(z=- 0.68\).

Step2: Calculate the P - value

For a two - tailed test, the P - value is \(P = 2\times(1 - \Phi(|z|))\), where \(\Phi\) is the cumulative distribution function of the standard normal distribution.
Since \(z=-0.68\), \(|z| = 0.68\). Using a standard normal table or a calculator with a normal distribution function (e.g., in Excel: \(=2*(1 - NORM.S.DIST(0.68,TRUE))\)), we find that \(P = 2\times(1 - 0.7517)=2\times0.2483 = 0.4966\approx0.497\)

Answer:

The P - value for this hypothesis test is \(0.497\)