?A team of researchers (Singer et al., 2000) used the Survey of Consumer Attitud

Question

?A team of researchers (Singer et al., 2000) used the Survey of Consumer Attitudes to investigate whether incentives would improve the response rates on telephone surveys. A national sample of 735 households was randomly selected, and all 735 of the households were sent an “advance letter” explaining that the household would be contacted shortly for a telephone survey. However, 368 households were randomly assigned to receive a monetary incentive along with the advance letter, and of these 286 responded to the telephone survey. The other 367 households were assigned to receive only the advance letter, and of these 245 responded to the telephone survey.

1. Use an appropriate randomization-based applet to find a p-value. Round your answer to two decimal places.

2. Based on this p-value, how much evidence do you have against the null hypothesis?

a. We have very strong evidence against the null hypothesis

b. We have no evidence against the null hypothesis

c. We have very weak evidence against the null hypothesis.

3. Calculate an appropriate standardized statistic in the context of the study. Round to two decimal places.

4. The p-value and standardized statistic both lead you to the same conclusion.

True or false?

5. Use the 2SD method to find a 95% confidence interval for the parameter of interest. Answers are supposed to be positive. Round your answers to two decimal places.

(_________,_________)

6.

Is your conclusion from using either the p-value or the standardized statistic consistent with your finding from the 95% confidence interval?

Yes or no?

Explanation / Answer

1) p1 = 286/368 = 0.777

    p2 = 245/367 = 0.668

The Pooled sample proportion (P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                                       = (0.777 * 368 + 0.668 * 367)/(368 + 367) = 0.723

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

      = sqrt(0.723 * (1 - 0.723) * (1/368 + 1/367))

      = 0.033

The test statstic z = (p1 - p2)/SE

                             = (0.777 - 0.668)/0.033 = 3.30

P-value = P(Z > 3.30)

             = 1 - P(Z < 3.30)

             = 1 - 0.9995

             = 0.0005

2) As the P-value is less than 0.05, so the null hypothesis is rejected.

Option - a) We have very strong evidence against the null hypothesis.

3) The test statstic z = (p1 - p2)/SE

                             = (0.777 - 0.668)/0.033 = 3.30

4) At 5% significance level the critical value is z* = 1.96

As the test statistic value is greater than the critical value (3.3 > 1.96), so the null hypothesis is rejected.

So its TRUE.

5) At 95% confidence interval the critical value is z0.025 = 1.96

The 95% confidence interval is

(p1 - p2) +/- z0.025 * SE

= 0.777 - 0.668 +/- 1.96 * 0.033

= 0.109 +/- 0.065

= 0.044, 0.174

= 0.04, 0.17

6) As the interval doesn't cointain 0, so the null hypothesis is rejected.Yes the conclusion is same

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