Suppose that in a random selection of 120 colored candies, 35% of them are red. The candy company claims that the percentage of red candies is equal to 40%. Use a 0.05 significance level to test that claim.

Identify the null and alternative hypotheses for this test.

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is ___.

Identify the P-value for this hypothesis test.
The P-value for this hypothesis test is ___.

Identify the conclusion for this hypothesis test.


Answer:
Given:

Hypothesized Population Proportion $ (p)=0.40 $

Sample Proportion $ (\hat{p})=0.35 $

Sample Size $ (n)=120 $

Significance Level $ (\alpha) = 0.05$

Solution:

The null and alternative hypothesis:

$H_0:p=0.40$
$H_a:p\neq0.40$

The test statistic:
$z = \frac{\hat{p} – p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.35 – 0.40}{\sqrt{\frac{0.40(1-0.40)}{120}}} = \frac{-0.05}{\sqrt{\frac{0.24}{120}}} = \frac{-0.05}{0.0447} = -1.12$

The p-value:
$\text{The p-value} = 2 \times P(z < -1.12) = 0.2636$

The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis, suggesting that there is not enough evidence to support the claim that the fraction of red candies is different from 40%.

Final answer:
The null and alternative hypothesis:
$H_0:p=0.40$
$H_a:p\neq0.40$

The test statistic $(z)=-1.12$

The p-value $=0.2636$

The conclusion:
We fail to reject the null hypothesis, suggesting that there is not enough evidence to support the claim that the fraction of red candies is different from 40%.

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