Suppose that in a random selection of 100 colored candies, of them are blue. The candy company claims that the percentage of blue candies is equal to . Use a 0.10 significance level to test that claim.

Identify the null and altemative hypotheses for this test. Choose the correct answer below.
A.

B.
C.
D. :

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is -1.31 .
(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 )

Question

Suppose that in a random selection of 100 colored candies, of them are blue. The candy company claims that the percentage of blue candies is equal to . Use a 0.10 significance level to test that claim.

Identify the null and altemative hypotheses for this test. Choose the correct answer below.
A.

B.
C.
D. :

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is -1.31 .
(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 )

Ask by Pope Bernard. in the United States

Mar 21,2025

UpStudy AI · Accepted Answer

**Step 1. State the Hypotheses**

The candy company claims that the percentage of blue candies is $$30\%$$. Thus, the null hypothesis is that the true proportion $$p$$ of blue candies is $$0.30$$. Since we want to test the claim (i.e., whether it is exactly $$30\%$$ or not), the alternative hypothesis is that the true proportion is not equal to $$0.30$$. This corresponds to a two-tailed test.

In mathematical terms, we have:

$$
H_{0}: p=0.3 \quad \text{and} \quad H_{1}: p \neq 0.3
$$

This matches option B.

---

**Step 2. Identify the Test Statistic**

The test statistic for a proportion is given by:

$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}}
$$

where:
- $$\hat{p}$$ is the sample proportion,
- $$p_0$$ is the claimed proportion (0.3),
- $$n$$ is the sample size.

Given:
- $$\hat{p} = 0.24$$ (since $$24\%$$ of 100 candies are blue),
- $$p_0 = 0.3$$,
- $$n = 100$$.

Plug the values into the formula:

$$
z = \frac{0.24 - 0.3}{\sqrt{\frac{0.3(1-0.3)}{100}}} = \frac{-0.06}{\sqrt{\frac{0.3 \times 0.7}{100}}} = \frac{-0.06}{\sqrt{0.21/100}} = \frac{-0.06}{\sqrt{0.0021}} = \frac{-0.06}{0.0458} \approx -1.31
$$

Thus, the test statistic is $$-1.31$$.

---

**Step 3. Find the P-value**

Since this is a two-tailed test, the p-value is calculated as:

$$
\text{P-value} = 2 \times P(Z < -1.31)
$$

Using a standard normal table or calculator, we find:

$$
P(Z < -1.31) \approx 0.095
$$

Thus,

$$
\text{P-value} \approx 2 \times 0.095 = 0.190
$$

---

**Final Answers**

1. **Hypotheses:**
   $$
   H_{0}: p=0.3 \quad \text{and} \quad H_{1}: p\neq 0.3 \quad \text{(Option B)}
   $$

2. **Test Statistic:**
   $$
   -1.31
   $$

3. **P-value:**
   $$
   0.190
   $$
**Hypotheses:** - Null Hypothesis ($$ H_{0} $$): The proportion of blue candies is 30% ($$ p = 0.3 $$) - Alternative Hypothesis ($$ H_{1} $$): The proportion of blue candies is not 30% ($$ p \neq 0.3 $$) **Test Statistic:** - $$-1.31$$ **P-value:** - $$0.190$$

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