Question

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

Answer 1

Answer:  a) -0.2252, b) 0.8219

Step-by-step explanation:

Since we have given that

Sample size n = 100

Probability that candies are blue = p= 0.26

Probability that company claims that it is blue candy = P = 0.27

So, Q = 1-P= 1-0.27 = 0.73

So, Null hypothesis : $H_0:p=P$

Alternate hypothesis : $H_1:p\neq P$

So, the test statistic would be

$z=\dfrac{p-P}{\sqrt{\dfrac{P.Q}{n}}}\\\\z=\dfrac{0.26-0.27}{\sqrt{\dfrac{0.27\times 0.73}{100}}}\\\\z=-0.2252$

Since α = 0.05

So, critical value of z = 1.96

p-value = P(Z>Z(calculated)

Using the excel function , we get that

$P(z>0.2252)\\\\=2\times 0.410.911845\\\\=0.8219$

Hence, a) -0.2252, b) 0.8219