QuestionDetermine if half of all teens aged 13-17 have made new friends online using a 0.05 significance level and the normal distribution.
$H_0: p = 0.5$ $H_1: p \neq 0.5$

Studdy Solution

STEP 1

Assumptions
1. The sample size is 1120 teens aged 13 to 17.
2. The sample proportion of teens who have made new friends online is 56%.
3. The significance level for the hypothesis test is 0.05.
4. The population proportion (denoted by $p$ ) is the proportion of all teens aged 13 to 17 who have made new friends online.
5. The null hypothesis ( $H_0$ ) will assume that the population proportion is equal to a specified value, which is typically the value under the claim that is being tested.
6. The alternative hypothesis ( $H_1$ ) will state that the population proportion is different from the value specified in the null hypothesis.
7. We will use the normal distribution to approximate the binomial distribution because the sample size is large.

STEP 2

Identify the null and alternative hypotheses. The null hypothesis will state that the population proportion is equal to 0.5 (the claim that half of all teens have made new friends online), and the alternative hypothesis will state that the population proportion is not equal to 0.5.
$H_0: p = 0.5 \\ H_1: p \neq 0.5$

STEP 3

Calculate the sample proportion ( $\hat{p}$ ) using the information given in the problem.
$\hat{p} = \frac{\text{Number of teens who have made new friends online}}{\text{Total number of teens surveyed}}$

STEP 4

Plug in the given values to calculate the sample proportion.
$\hat{p} = \frac{56\% \times 1120}{1120}$

STEP 5

Convert the percentage to a decimal and simplify the sample proportion.
$\hat{p} = \frac{0.56 \times 1120}{1120} = 0.56$

STEP 6

Calculate the standard error (SE) of the sampling distribution using the null hypothesis proportion ( $p_0 = 0.5$ ) and the sample size (n = 1120).
$SE = \sqrt{\frac{p_0(1 - p_0)}{n}}$

STEP 7

Plug in the values for $p_0$ and $n$ to calculate the standard error.
$SE = \sqrt{\frac{0.5(1 - 0.5)}{1120}}$

STEP 8

Simplify the standard error calculation.
$SE = \sqrt{\frac{0.25}{1120}} = \sqrt{0.000223214} \approx 0.01493$

STEP 9

Calculate the test statistic (z-score) using the sample proportion ( $\hat{p}$ ), the null hypothesis proportion ( $p_0$ ), and the standard error (SE).
$z = \frac{\hat{p} - p_0}{SE}$

STEP 10

Plug in the values for $\hat{p}$ , $p_0$ , and SE to calculate the test statistic.
$z = \frac{0.56 - 0.5}{0.01493}$

STEP 11

Simplify the calculation to find the z-score.
$z = \frac{0.06}{0.01493} \approx 4.02$

STEP 12

Use the z-score to find the P-value. Since this is a two-tailed test (because the alternative hypothesis is $p \neq 0.5$ ), we need to find the probability that a standard normal variable is more extreme than the absolute value of our test statistic.
$P\text{-value} = 2 \times P(Z > |z|)$

STEP 13

Look up the P-value corresponding to the z-score of 4.02 in the standard normal distribution table or use a calculator with normal distribution functions.
$P\text{-value} \approx 2 \times (1 - 0.9999) = 2 \times 0.0001 = 0.0002$

STEP 14

Compare the P-value to the significance level ( $\alpha = 0.05$ ). If the P-value is less than the significance level, we reject the null hypothesis.
$P\text{-value} < \alpha$
$0.0002 < 0.05$

STEP 15

Since the P-value is less than the significance level, we reject the null hypothesis.
The conclusion is that there is sufficient evidence at the 0.05 significance level to reject the claim that half of all teens have made new friends online.
The values to fill in the table are:
$H_0: P = 0.5 \\ H_1: P \neq 0.5$

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