Question

Suppose that in a random selection of 100 colored candies, 23% of them are blue. The candy company claims that the percentage of blue candies is equal to 28%. Use a .10

significance level to test that claim. Identify the test statistic and the P value for this claim.

Answer 1

Solution :

Given that,

This a two- tailed test.

The null and alternative hypothesis is,

Ho: p = 0.28

Ha: p 0.28

Test statistics

z = ( - ) / *(1-) / n

= ( 0.23 - 0.28) / (0.23*0.77) /100

= -1.188

P-value = 2 * P(Z < z )

= 2 * P(Z < -1.188 )

= 2 * 0.1174

= 0.2348

The p-value is p = 0.2348, and since p = 0.2348 > 0.10, it is concluded that the null hypothesis is fail to rejected.

There is not sufficient evidence to claim that the percentage of blue candies is equal to 28%.