Question

(Question) Web Check response rates. Response rates to Web checks are generally low, incompletely due to druggies starting but not. I am finishing the check. Survey Methodology (December 2013) delved into the factors that impact response rates. In a designed study, Web druggies were directed to. Share in one of several checks with different formats. For illustration, one format employed a welcome screen with a white background, and another format employed a welcome screen with a red background. The “break-off rates,” i.e., the proportion of tried druggies who break off the check before completing all questions, for the two formats are handed in the table.

	White Welcome screen	Red Welcome screen
Number of Web users	198	183
The number who break off the survey	49	37
Break-off rate	.258	.202

Source: R. Haer and N. Meidert, “Does the First Impression Count? Examining the Effect of the Welcome Screen Design on the Response Rate,” Survey Methodology, Vol. 39, No. 2, December 2013 (Table 4.1).

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

b. The experimenters theorize that the true break-off rate for Web druggies of the red hello screen will be lower than the corresponding break-off rate for the white hello screen. Give the null and indispensable suppositions for testing this proposition.

c. Cipher the test statistic for the test.

d. Find the p- the value of the test.

e. Make the applicable conclusion using α = .10.

Short answer

A test statistic assesses the level of agreement among a data set and the null hypothesis. Its recorded values ranged randomly from one random test to another.

Step-by-step solution

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

Consider X_red= 49 and n_eed= 190.

The value of the break-off rate for the white welcome screen is,

$$\begin{gathered} P_{red}=\frac{x_{white}}{n_{white}} \\ =\frac{49}{130} \\ =0.258 \end{gathered}$$

Consider x_red = 37 and n_red= 183.

The value of the break-off rate for the red welcome screen is,

$$\begin{gathered} P_{red}=\frac{x_{white}}{n_{white}} \\ =\frac{37}{183} \\ =0.202 \end{gathered}$$

The value of the break-off rates shown in the table is verified.

State the hypotheses

Null hypothesis:

H₀ = P_white - P_red= 0

The real break-off rate for Internet users who see the red welcome display will be smaller than the actual break-off percentage for Internet users who see the white welcome display.

Alternative hypothesis:

H₀: W_hite-P_red > 0

The real break-off rate for Internet users who see the red welcome display will be smaller than the actual break-off percentage for Internet users who see the white welcome display.

Use MINITAB to obtain the test statistic

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 2 Proportions.

Step 2: Choose Summarized data.

Step 3: In the First sample, enter Trials as 190 and Events as 49.

Step 4 in the Second sample, enter Trials as 183 and Events as 37.

Step 5: Check Perform hypothesis test. In Hypothesized proportion, enter 0.

Step 6: Check Options, enter Confidence level as 90.0

Step 7: Choose greater than in alternative

Step 8: Click OK in all dialogue boxes.

MINITAB output

Test and CI for Two Proportions

Difference = p (1) - p (2)

Estimate for difference: 0.0557089

90% lower bound for difference: 0.0000130753

Test for difference = 0 (vs > 0) : z = 1.28 P-Value = 0.101

Fisher's exact test - P-Value 0.124

From the MINITAB output, the value of the test statistic is 128.

Value of p

Finds the p-value by using MINITAB.

From the MINITAB output in step-4, the value is 0.101.

Rejection rule and conclusion

Rejection rule:

If p-value < α . then reject the null hypothesis.

Conclusion:

Here, the p-value is greater than the level of significance.

That is p-value (-0,101) > α(-0,05)

Therefore, the null hypothesis is not rejected at α = 0.05

It can be concluded that the actual break-off rate for Web users of the red welcome screen is hot lower than the corresponding true break-off rate for the white welcome screen.