Use a 0.05 Significance level to test the claim that 20% of all skittles candies are red. 243 skittles were red out of 1245 skittles.

We are told that 243 of 1245 skittles(tm) are red, and we are asked to test the claim that 20% of skittles are red at the .05 significance level.

(1) The null hypothesis is H0: 20% of skittles are red. (The claim)

The alternative hypothesis H1: The true percentage is not 20%.

(2) This is a two-tailed test. Since we are dealing with proportions we use a z-test to find the critical value(s) and the...

We are told that 243 of 1245 skittles(tm) are red, and we are asked to test the claim that 20% of skittles are red at the .05 significance level.

(1) The null hypothesis is H0: 20% of skittles are red. (The claim)

The alternative hypothesis H1: The true percentage is not 20%.

(2) This is a two-tailed test. Since we are dealing with proportions we use a z-test to find the critical value(s) and the critical region(s):

For alpha = .05 we have critical values of -1.96 and 1.96. (These are from a standard normal table; the area to the left of z=-1.96 is approximately .025, while the area to the right of z=1.96 is approximately .025.)

(3) To compute the test value we take the observed value  (approximately .1952) minus the expected value (.2); divide this by the standard error which is given by the square root of the product of p (.2), 1-p (.8) divided by n=1245. Thus the test value is -.0048/.0113 or z=-.425

(4) The test value is not in the critical region (-1.96<-.425<1.96 ) so we do not reject the null hypothesis.

(5) There is insufficient evidence to reject the claim that 20% of skittles are red.

** My calculator gives the p-value as .6707; since this is greater than alpha we do not reject the null hypothesis as above.