A poll of 1168 teens aged 13 to 17 showed that $$ 53 \% $$ of them have made new friends online. Use a 0.01 significance level to test the claim that half of all teens have made new friends online. Use the P-val method. Use the normal distribution as an approximation to the binomial distribution. 0 of 4 Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online Identify the null and altemative hypotheses. $$ \mathrm{H}_{0} \cdot \mathrm{p}=0.5 $$ $$ \mathrm{H}_{1} \mathrm{p}
eq 0.5 $$ (Type integers or decimals. Do not round.) Identify the test statistic. $$ \mathrm{z}=2.05 $$ (Round to two decimal places as needed.) Identify the P-value. P-value $$ =\square $$ (Round to three decimal places as needed.) a

Question

A poll of 1168 teens aged 13 to 17 showed that $$ 53 \% $$ of them have made new friends online. Use a 0.01 significance level to test the claim that half of all teens have made new friends online. Use the P-val method. Use the normal distribution as an approximation to the binomial distribution. 0 of 4 Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online Identify the null and altemative hypotheses. $$ \mathrm{H}_{0} \cdot \mathrm{p}=0.5 $$ $$ \mathrm{H}_{1} \mathrm{p} 
eq 0.5 $$ (Type integers or decimals. Do not round.) Identify the test statistic. $$ \mathrm{z}=2.05 $$ (Round to two decimal places as needed.) Identify the P-value. P-value $$ =\square $$ (Round to three decimal places as needed.) a

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**Step 1. State the Hypotheses**

We are testing the claim that half of all teens have made new friends online.

- Null hypothesis: $$ H_0: p = 0.5 $$
- Alternative hypothesis: $$ H_1: p \neq 0.5 $$

---

**Step 2. Compute 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

- $$\hat{p} = 0.53$$ is the sample proportion,
- $$p_0 = 0.5$$ is the hypothesized proportion, and
- $$n = 1168$$ is the sample size.

Calculate the standard error:

$$
\text{SE} = \sqrt{\frac{0.5 \times 0.5}{1168}} = \sqrt{\frac{0.25}{1168}} \approx \sqrt{0.000214} \approx 0.01464.
$$

Now, compute $$z$$:

$$
z = \frac{0.53 - 0.5}{0.01464} \approx \frac{0.03}{0.01464} \approx 2.05.
$$

---

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

Since $$ H_1: p \neq 0.5 $$ is a two-tailed test, the P-value is given by

$$
\text{P-value} = 2 \Pr(Z > |2.05|).
$$

Looking up the normal probability for $$z = 2.05$$, we have

$$
\Pr(Z > 2.05) \approx 0.0202.
$$

Thus, the two-tailed P-value is

$$
\text{P-value} = 2 \times 0.0202 \approx 0.0404.
$$

Rounded to three decimal places, the P-value is

$$
0.040.
$$

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**Final Answers**

- Null and alternative hypotheses:
  - $$ H_0: p = 0.5 $$
  - $$ H_1: p \neq 0.5 $$
- Test statistic: $$ z = 2.05 $$
- P-value: $$ 0.040 $$
**Hypotheses:** - $$ H_0: p = 0.5 $$ - $$ H_1: p \neq 0.5 $$ **Test Statistic:** $$ z = 2.05 $$ **P-value:** 0.040

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