Question

Reducing nonresponse? Telephone surveys often have high rates of nonresponse. When the call is handled by voice mail, leaving a message might encourage people to respond when they are called again. Here are data form a study in which (at random) a message was or was not left when voice mail picked up the first call from a survey:

(a) Is this an experiment or an observational study? Justify your answer.

(b) Based on the data, can we conclude that leaving a message increases the proportion of households that are eventually interviewed? Give appropriate statistical evidence to support your answer

Short answer

a). It is an experiment.

b). Yes, we can conclude that leaving a message increases the proportion of households that are eventually interviewed.

Step-by-step solution

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

An experiment intentionally applies a certain procedure on people to study their reactions. An observational analysis aims to gain details without disrupting the scene they are watching.

Part (b) Step 1: Given Information

$x_1=200$

$n_1=200+91$

$=291$

$x_2=58$

$n_2=58+42$

$=100$

Part (b) Step 2: Explanation

The hypotheses are

$H_0:p_1=p_2$

$H_0:p_1>p_2$

The sample proportion is

${\hat{p}}_1=\frac{x_1}{n_1}$

$=\frac{200}{291}$

$=0.6873$

${\hat{p}}_2=\frac{x_2}{n_2}$

$=\frac{58}{100}$

$=0.58$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}$

$=\frac{200+58}{291+100}$

$=0.6598$

Part (b) Step 3: Explanation

The test statistic is

$z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$=\frac{0.6873-0.58}{\sqrt{0.6598(1-0.6598)\sqrt{\frac{1}{291}+\frac{1}{100}}}}$

$=1.95$

The P-value is the chance of having the value of the test numbers, or a more drastic value.

$P=P(Z>1.95)$

$=P(Z<-1.95)$

$=0.0256$

Reject the null hypothesis if the P-value is lower than the significance level:

localid="1650389579178" $P<0.05\Rightarrow \text{Reject}{H}_0$