In: Statistics and Probability
Suppose that in a random selection of 100 colored candies, 25% of them are blue. The candy company claims that the percentage of blue candies is equal to 21%. Use a 0.10 significance level to test that claim. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0: pequals0.21 Upper H 1: pgreater than0.21 B. Upper H 0: pnot equals0.21 Upper H 1: pequals0.21 C. Upper H 0: pequals0.21 Upper H 1: pless than0.21 D. Upper H 0: pequals0.21 Upper H 1: pnot equals0.21 Your answer is correct. Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is nothing. (Round to two decimal places as needed.).
The correct null and alternative hypotheses for this test is as below
D. Upper H 0: pequals0.21 Upper H 1: pnot equals0.21
By using parameter P we also write the above null and alternative hypothesis
That is,the proportion of blue candies is equal to 0.21.
That is,the proportion of blue candies is not equal to 0.21.
The values provided in the question are as below
The total sample selection = n = 100
sample proportion = = 25% = 0.25
population proportion = = 0.21
The formula of test statistic is
The test statistic = 0.98
We find the critical values using 0.10 significance level
The above test is two tailed test
Hence, cumulative area to the left is
Area to the left = 1 - Area to the right
= 1 - 0.05
= 0.95
Using Excel function =NORMSINV(probability)
= NORMSINV(0.95)
The test is two tailed so, we can write z value using sign.
[EXCEL formula = NORMSINV(0.95)]
Critical value of /2 = 1.64
We comparing the test statistic with Critical value of /2 and take decision about the accept or reject the null hypothesis.
The test statistic = 0.98 not lies in the Critical value of /2 = 1.64
We accept Ho. the claim is true for the proportion of blue candies is equal to 0.21.