I believe the population proportion of U.S. children ages 8 to 18 years old who had internet access at home is 0.84 or 84% as of January 2010. A randomly selected survey of 500 children ages 8 to 18 years old was taken, and the results showed that 430 of them had internet access at home. Do the results of this survey suggest that the proportion of U.S. children who had internet access has changed since 2010? We will use a significance level of 0.05.
Null Hypothesis: p = 0.84
Alternative Hypothesis: p ≠ 0.84
We will use the p-value method to test the hypothesis. The p-value is the probability of obtaining a sample proportion as extreme as the one observed, assuming the null hypothesis is true.
To calculate the p-value, we will use the normal approximation to the binomial distribution since the sample size is large (n = 500) and the conditions for using the normal approximation are met.
The test statistic can be calculated using the formula:
z = (p̂ - p) / √(p * (1 - p) / n)
where p̂ is the sample proportion, p is the hypothesized population proportion, and n is the sample size.
In this case, p̂ = 430/500 = 0.86, p = 0.84, and n = 500.
Plugging in these values, we get:
z = (0.86 - 0.84) / √(0.84 * (1 - 0.84) / 500)
z = 0.02 / √(0.84 * 0.16 / 500)
z = 0.02 / √(0.1344 / 500)
z = 0.02 / √0.0002688
z = 0.02 / 0.0164
z ≈ 1.22
To find the p-value, we will use a standard normal distribution table or a calculator. The p-value is the probability of obtaining a z-score as extreme as 1.22 or more extreme in either direction.
Based on the standard normal distribution table, the p-value is approximately 0.2236.
Since the p-value (0.2236) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to suggest that the proportion of U.S. children who had internet access has changed since 2010.