Sometimes parents and grandparents like to recount how difficult life was when they were kids, such as having to walk 10+ miles to school (in the snow, uphill both ways). A random sample of 180 teenagers were selected and 40% had heard stories from their parents or grandparents about how difficult life was when they were kids. Do these data provide convincing evidence at an α = 0.05 significance level that the proportion of all teenagers who have heard stories from their parents or grandparents about how difficult life was when they were kids differs from 0.50? State: Plan: Do: Conclude:

Answer:

Sometimes parents and grandparents like to recount how difficult life was when they were kids, such as having to walk 10+ miles to school (in the snow, uphill both ways). A random sample of 180 teenagers were selected and 40% had heard stories from their parents or grandparents about how difficult life was when they were kids. Do these data provide convincing evidence at an α = 0.05 significance level that the proportion of all teenagers who have heard stories from their parents or grandparents about how difficult life was when they were kids differs from 0.50? State: Plan: Do: Conclude:

Explanation:

We want to test if the proportion of all teenagers who have heard stories from their parents or grandparents about how difficult life was when they were kids differs from 0.50.

The significance level is α = 0.05.

The sample size is n = 180 and the proportion of the sample who have heard stories is p = 0.40.

Plan:

We will use a one-sample proportion test.

The null hypothesis is that the true proportion, p, is equal to 0.50.

The alternative hypothesis is that the true proportion, p, is not equal to 0.50.

We will use a significance level of α = 0.05.

Do:

The test statistic is z = (p - P) / sqrt(P * (1 - P) / n), where P is the hypothesized proportion under the null hypothesis.

The observed test statistic is z = (0.40 - 0.50) / sqrt(0.50 * (1 - 0.50) / 180) = -2.47 (rounded to two decimal places).

The p-value is P(Z ≤ -2.47) + P(Z ≥ 2.47) = 2 * P(Z ≤ -2.47) = 0.013 (rounded to three decimal places), where Z is the standard normal distribution.

Since the p-value is less than the significance level of α = 0.05, we reject the null hypothesis.

Conclude:

There is convincing evidence at an α = 0.05 significance level that the proportion of all teenagers who have heard stories from their parents or grandparents about how difficult life was when they were kids differs from 0.50.

Specifically, the sample proportion of 0.40 suggests that the proportion may be lower than 0.50.

We want to test if the proportion of all teenagers who have heard stories from their parents or grandparents about how difficult life was when they were kids differs from 0.50. The significance level is a 0.05.

The sample size is n 180 and the proportion of the sample who have heard stories is p=0.40. Plan: We will use a one-sample proportion test. The null hypothesis is that the true proportion, p, is equal to 0.50. The alternative hypothesis is that the true proportion, p. is not equal to 0.50 We will use a significance level of a=0.05.

Do: The test statistic is z=(p-P)/sqrt(P(1-P)/n), where P is the hypothesized proportion under the null hypothesis. The observed test statistic is z = (0.40-0.50)/sqrt(0.50 (1-0.50)/180) = -2.47 (rounded to two decimal places). The p-value is P(Z<-247) + P(Z ≥2.47)-2 PZ ≤ 2.47) 0.013 (rounded to three decimal places), where Z is the standard normal distribution. Since the p-value is less than the significance level of a=0.05, we reject the null hypothesis. Conclude: There is convincing evidence at an a = 0.05 significance level that the proportion of all teenagers who have heard stories from their parents or grandparents about how difficult life was when they were kids differs from 0.50.

Specifically, the sample proportion of 0.40 suggests that the proportion may be lower than 0.50.

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