Question

suppose that in a random selection of 100 colored candies, 21% of them are blue. the candy company claims that the percentage of blue candies is equal to 27%. use a 0.01 significance level to test that claim.

identify the null and alternative hypotheses for this test. choose the correct answer below

oa. ( h_0: p = 0.27 )
( h_1: p>0.27 )

ob. ( h_0: p
eq0.27 )
( h_1: p = 0.27 )

oc. ( h_0: p = 0.27 )
( h_1: p<0.27 )

od. ( h_0: p = 0.27 )
( h_1: p
eq0.27 )

identify the test statistic for this hypothesis test.

the test statistic for this hypothesis test is
(round to two decimal places as needed.)

identify the p - value for this hypothesis test.

the p - value for this hypothesis test is
(round to three decimal places as needed.)

identify the conclusion for this hypothesis test.

oa. fail to reject ( h_0 ). there is not sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 27%

ob. fail to reject ( h_0 ). there is sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 27%

oc. reject ( h_0 ). there is sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 27%

od. reject ( h_0 ). there is not sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 27%

Explanation:

Step1: Identify null and alternative hypotheses

The null hypothesis \(H_0\) is the claim being tested. Here, the claim is \(p = 0.27\). The alternative hypothesis \(H_1\) for a two - tailed test (since we are just testing if the proportion is different) is \(p
eq0.27\). So \(H_0:p = 0.27\) and \(H_1:p
eq0.27\) (Option D).

Step2: Calculate the test statistic

The formula for the test statistic \(z\) in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Given \(\hat{p}=0.21\), \(p = 0.27\), \(n = 100\)

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Step3: Calculate the P - value

For a two - tailed test with \(z=-1.35\), the P - value is \(2P(Z\lt - 1.35)\)
Using a standard normal table or calculator, \(P(Z\lt - 1.35)=0.0885\)
So \(P - value=2\times0.0885 = 0.177\)

Step4: Make a conclusion

Since the P - value (\(0.177\))\(>0.01\) (significance level \(\alpha\)), we fail to reject \(H_0\). There is not sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to \(27\%\) (Option A)

Answer:

• Hypotheses: D. \(H_0:p = 0.27\), \(H_1:p

eq0.27\)

• Test statistic: \(-1.35\)
• P - value: \(0.177\)
• Conclusion: A. Fail to reject \(H_0\). There is not sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to \(27\%\)