Question

An organization surveyed 1000 teens and 1000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace. The two samples were independently selected and were chosen in a way that makes it reasonable to regard them as representative of American teens and parents of American teens.

(a) When asked if they check their online social networking sites more than 10 times a day, 226 of the teens surveyed said yes. When parents of teens were asked if their teen checked his or her site more than 10 times a day, 49 said yes. Use a significance level of 0.01 to carry out a hypothesis test to determine if there is convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day. (Use μ_teens − μ_parents. Round your test statistic to two decimal places and your P-value to four decimal places.)

z= ____ P-value= ____

(b) The article also reported that 381 of the teens surveyed said they had posted something on their networking site that they later regretted. Would you use the two-sample z test of this section to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted?

(c) Using an appropriate test procedure, carry out a test of the hypothesis given in part (b). Use α = 0.05 for this test. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z= ____  P-value= ____

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P_Teen< P_Parents
Alternative hypothesis: P_Teen > P_Parents

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.1375

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.015401
z = (p₁ - p₂) / SE

z = 11.49

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 11.49.

Thus, the P-value = less than 0.0001 (0.0000)

Interpret results. Since the P-value (almost 0) is less than the significance level (0.01), we failed to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day.

b) We would use one -sample z test to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted.

c)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.333
Alternative hypothesis: P > 0.333

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.014907
z = (p - P) / S.D

z = 3.22

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 3.22.

Thus, the P-value = 0.0006

Interpret results. Since the P-value (0.0006) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that more than one-third of all teens have posted something on a social networking site that they later regretted.