Question

Suppose that in a random selection of 100 colored​ candies, 24​% of them are blue. The candy company claims that the percentage of blue candies is equal to 23​%. Use a 0.01 significance level to test that claim.

Answer 1

Answer:

We conclude that the percentage of blue candies is equal to 23​% at 0.01 significance level.

Step-by-step explanation:

We are given that in a random selection of 100 colored​ candies, 24​% of them are blue.

The candy company claims that the percentage of blue candies is equal to 23​%.

Let p = percentage of blue candies.

So, Null Hypothesis, $H_0$ : p = 23%      {means that the percentage of blue candies is equal to 23​%}

Alternate Hypothesis, $H_A$ : p $\neq$ 23%      {means that the percentage of blue candies is different from 23​%}

The test statistics that would be used here One-sample z proportion test statistics;

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

where, $\hat p$ = sample proportion of blue colored candies = 24%

           n = sample of colored​ candies = 100

So, test statistics  =  $\frac{0.24-0.23}{\sqrt{\frac{0.24(1-0.24)}{100}} }$

                               =  0.234

The value of z test statistics is 0.234.

Now, at 0.01 significance level the z table gives critical values of -2.5758 and 2.5758 for two-tailed test.

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

Therefore, we conclude that the percentage of blue candies is equal to 23​%.