In: Math
Truthfulness in online profiles. Many teens have posted profiles
on a social-
networking website. A sample survey in 2007 asked a random sample
of teens
with online profiles if they included false information in their
profiles. Of 170
younger teens (aged 12 to 14), 117 said yes. Of 317 older teens
(aged 15 to 17),
152 said yes.
16
(a) Do these samples satisfy the guidelines for the large-sample
confidence
interval?
(b) Give a 95% confidence interval for the difference between the
proportions
of younger and older teens who include false information in their
online
profiles.
(a) Do these samples satisfy the guidelines for the large-sample confidence interval?
Answer: Yes, these samples satisfy the guidelines for the large-sample confidence interval because both samples have sample sizes greater than 30, and both samples are random samples.
(b) Give a 95% confidence interval for the difference between
the proportions
of younger and older teens who include false information in their
online
profiles.
Confidence interval for difference between two population proportions:
Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]
Where, P1 and P2 are sample proportions for first and second groups respectively.
We are given
X1 = 117
N1 = 170
P1 = X1/N1 = 117/170 = 0.688235294
X2 = 152
N2 = 317
P2 = X2/N2 = 152/317 = 0.479495268
Confidence level = 95%
Critical Z value = 1.96
(by using z-table)
(P1 – P2) = 0.688235294 - 0.479495268 = 0.208740026
Standard error = sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]
Standard error = sqrt[(0.688235294*(1 – 0.688235294)/170) + (0.479495268*(1 – 0.479495268)/317)]
Standard error = 0.0453
Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]
Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]
Confidence interval = 0.208740026 ± 1.96*0.0453
Confidence interval = 0.208740026 ± 0.0887
Lower limit = 0.208740026 - 0.0887 = 0.1200
Upper limit = 0.208740026 + 0.0887 = 0.2975
Confidence interval = (0.1200, 0.2975)