Suppose that in a random selection of 100 colored candies, 26% 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. Identify the test statistic for this hypothesis test. find the P-value for this hypothesis test
Accepted Solution
A:
Answer: a) -0.2252, b) 0.8219Step-by-step explanation:Since we have given that Sample size n = 100Probability that candies are blue = p= 0.26Probability that company claims that it is blue candy = P = 0.27So, Q = 1-P= 1-0.27 = 0.73So, Null hypothesis : [tex]H_0:p=P[/tex]Alternate hypothesis : [tex]H_1:p\neq P[/tex]So, the test statistic would be [tex]z=\dfrac{p-P}{\sqrt{\dfrac{P.Q}{n}}}\\\\z=\dfrac{0.26-0.27}{\sqrt{\dfrac{0.27\times 0.73}{100}}}\\\\z=-0.2252[/tex]Since α = 0.05So, critical value of z = 1.96p-value = P(Z>Z(calculated)Using the excel function , we get that [tex]P(z>0.2252)\\\\=2\times 0.410.911845\\\\=0.8219[/tex]Hence, a) -0.2252, b) 0.8219