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:
[tex]H_{0}: p = 0.29[/tex]
We want to test the hypothesis that this is true, so the alternate hypothesis is:
[tex]H_{a}: p \neq 0.29[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.29 is tested at the null hypothesis:
This means that [tex]\mu = 0.29, \sigma = \sqrt{0.29*0.71}[/tex]
Suppose that in a random selection of 100 colored candies, 21% of them are blue.
This means that [tex]n = 100, X = 0.21[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}[/tex]
[tex]z = -1.76[/tex]
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%.
