A random sample of 100 males found that 80% watched a particular TV show in the past week. A random sample of 100 females found that 50% watched the...

A random sample of 100 males found that 80% watched a particular TV show in the past week. A random sample of 100 females found that 50% watched the same TV show in the past week.

The standard error for the sampling distribution of the sample proportion for males is:

0.0016

0.04

0.08

0.16

The standard error for the sampling distribution of the sample proportion for females is:

0.25

0.0025

0.10

0.05

What is the standard error for the difference between our two sample values?

√(0.04 + .05) = √0.09 = 0.30

√((0.04)^2 + (0.05)^2) = √0.0041 = 0.064

√((0.0016)^2 + (0.0025)^2) = √0.000009 = 0.003

√(0.08 + 0.10) = √0.18 = .42

What is the formula for a 95% confidence interval of the difference between the two population proportions?

(sample proportion 1 - sample proportion 2) +/- 1*SED

(sample proportion 1 - sample proportion 2) +/- 3*SED

(sample proportion 1 - sample proportion 2) +/- 2

(sample proportion 1 - sample proportion 2) +/- 2 *SED

What is the 95% confidence interval of the difference between the two population proportions?

(.80 - .50) +/- 1 * 0.003 = .30 +/- 0.003 = (0.297 to 0.303) = (29.7% to 30.3%)

(.80 - .50) +/- 2 * 0.003 = .30 +/- 0.006 = (0.294 to 0.306) = (29.4% to 30.6%)

(.80 - .50) +/- 3 * 0.003 = .30 +/- 0.009 = (0.291 to 0.309) = (29.1% to 30.9%)

(.80 - .50) +/- 1 * 0.064 = .30 +/- 0.064 = (0.236 to 0.364) = (23.6% to 36.4%)

(.80 - .50) +/- 2 * 0.064 = .30 +/- 0.128 = (0.172 to 0.428) = (17.2% to 42.8%)

(.80 - .50) +/- 3 *0.064 = .30 +/- 0.192 = (0.108 to 0.492) = (10.8% to 49.2%)

The confidence interval in #5 shows that:

the difference between the population proportions for the two groups is significant

the difference between the population proportions for the two groups is not significant

cannot tell from the information given