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)=0.35 $
Sample Proportion $ (\hat{p})=0.32 $
Sample Size $ (n)=200 $
Significance Level $ (\alpha) = 0.05$
Solution:
The null and alternative hypothesis:
$H_0:p=0.35$
$H_a:p\neq0.35$
The test statistic:
$z = \frac{\hat{p} – p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.32 – 0.35}{\sqrt{\frac{0.35(1-0.35)}{200}}} = \frac{-0.03}{\sqrt{\frac{0.2275}{200}}} = \frac{-0.03}{0.0337} = -0.89$
The p-value:
$\text{The p-value} = 2 \times P(z < -0.89) = 0.3737$
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 orange candies is different from 35%.
Final answer:
The null and alternative hypothesis:
$H_0:p=0.35$
$H_a:p\neq0.35$
The test statistic $(z)=-0.89$
The p-value $=0.3737$
The conclusion:
We fail to reject the null hypothesis, suggesting that there is not enough evidence to support the claim that the fraction of orange candies is different from 35%.
