Response rates to Web surveys are typically low, partially due to users starting but not finishing the survey.

Question

Response rates to Web surveys are typically low, partially due to users starting but not finishing the survey. The factors that influence response rates were investigated in Survey Methodology (Dec. 2013). In a designed study, Web users were directed to participate in one of several surveys with different formats. For example, one format utilized a welcome screen witb a white background, and another format utilized a welcome screen with a red background. The "break-off rates," i.e., the proportion of sampled users who break off the survey before completing all questions, for the two formats are provided in the table.

$\begin{array}{cc} White Welcome Screen & Red Welcome Screen \\ Number of Web users & 190 & 183 \\ Number who break off survey & 49 & 37 \\ Break-off rate & 0.258 & 0.202 \end{array}$

a) Verify the values of the break-off rates shown in the table.

b) The researchers theorize that the true break-off rate for Web users of the red welcome screen will be lower than the corresponding break-off Tate for users of the white welcome screen. Give the null and alternative hypotbesis for testing this theory.

c) Conduct the test, part b, at $α = 0.10 .$ What do you conclude?

Answer 1

Step 1

Given: $n_{1} = Sample size = 190$
$n_{2} = Sample size = 183$
$x_{1} = Number of successes = 49$
$x_{2} = Number of successes = 37$
${\hat{p}}_{1} = Sample proportion = 0.258$
${\hat{p}}_{2} = Sample proportion = 0.202$
$α = Significance level = 0.10$ a) The sample proportion is the number of successes divided by the sample size: ${\hat{p}}_{1} = \frac{x_{1}}{n_{1}} = \frac{49}{190} \approx 0.258$
${\hat{p}}_{2} = \frac{x_{2}}{n_{2}} = \frac{37}{183} \approx 0.202$

Step 2

b) Given claim: Proportion is lower for red welcome screen. The claim iseither the null hypotethesis or the alternative hypothesis. The null hypothesis states that the population proportions areequal. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis. $H_{0} : p_{1} = p_{2}$
$H_{α} : p_{1} > p_{2}$

Step 3

c) The sample proportion is the number of successed divided by the sample size: $\hat{p_{1}} = \frac{x_{1}}{n_{1}} = \frac{49}{190} \approx 0.258$
$\hat{p_{2}} = \frac{x_{2}}{n_{2}} = \frac{37}{183} \approx 0.202$
$\hat{p} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}} = \frac{49 + 37}{190 + 183} = \frac{86}{373} \approx 0.2306$ Determine the value of the test statistic: $z = \frac{{\hat{p}}_{1} - {\hat{p}}_{2} - (p_{1} - p_{2})}{\sqrt{\hat{p} (1 - \hat{p})} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} = \frac{0.258 - 0.202 - 0}{\sqrt{0.2306 (1 - 0.2306} \sqrt{\frac{1}{190} + \frac{1}{183}}} \approx 1.28}$ Step 4 If the alternative hypothesis

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Response rates to Web surveys are typically low, partially due to users starting but not finishing the survey.

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