Question

Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%. Use the 0.01 significance level to test that claim.

Answer 1

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

$H_{0}: p = 0.29$

We want to test the hypothesis that this is true, so the alternate hypothesis is:

$H_{a}: p \neq 0.29$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that $\mu = 0.29, \sigma = \sqrt{0.29*0.71}$

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that $n = 100, X = 0.21$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}$

$z = -1.76$

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.