Suppose that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%. Use a 0.10 significance level to test that claim.
A. What is the test statistic for the hypothesis test?
B. What is the p value?
C. Reject/fail to reject sufficient evidence.

We are given that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%.

Let p = population percentage of blue candies

So, Null Hypothesis, [tex]H_0[/tex] : p = 29% {means that the percentage of blue candies is equal to 29%}

Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 29% {means that the percentage of blue candies is different from 29%}

The test statistics that will be used here is One-sample z-test for proportions;

T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of blue coloured candies = 28%

n = sample of colored candies = 100

So, the test statistics = [tex]\frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }[/tex]

= -0.22

The value of the z-test statistics is -0.22.

Also, the P-value of the test statistics is given by;

P-value = P(Z < -0.22) = 1 - P(Z [tex]\leq[/tex] 0.22)

= 1 - 0.5871 = 0.4129

Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, so we insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29%.