(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

Expert verified

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

01

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

Consider Xred= 49 and need= 190.

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

Pred=xwhitenwhite=49130=0.258

Consider xred = 37 and nred= 183.

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

Pred=xwhitenwhite=37183=0.202

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

02

State the hypotheses

Null hypothesis:

H0 = Pwhite - Pred= 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: White-Pred > 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.

03

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.

04

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.

05

Value of p

Finds the p-value by using MINITAB.

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

06

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.

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Most popular questions from this chapter

Solar energy generation along highways. Refer to the International Journal of Energy and Environmental Engineering (December 2013) study of solar energy generation along highways, Exercise 8.39 (p. 481). Recall that the researchers compared the mean monthly amount of solar energy generated by east-west– and north-south– oriented solar panels using a matched-pairs experiment. However, a small sample of only five months was used for the analysis. How many more months would need to be selected to estimate the difference in means to within 25 kilowatt-hours with a 90% confidence interval? Use the information provided in the SOLAR file to find an estimate of the standard error required to carry out the calculation

Question: Deferred tax allowance study. A study was conducted to identify accounting choice variables that influence a manager’s decision to change the level of the deferred tax asset allowance at the firm (The Engineering Economist, January/February 2004). Data were collected for a sample of 329 firms that reported deferred tax assets in 2000. The dependent variable of interest (DTVA) is measured as the change in the deferred tax asset valuation allowance divided by the deferred tax asset. The independent variables used as predictors of DTVA are listed as follows:

LEVERAGE: x1= ratio of debt book value to shareholder’s equity

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0 if not

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BBATH: x4 = 1 if operating earnings negative and lower than last year,

0 if not

EARN: x5 = change in operating earnings divided by total assets

A first-order model was fit to the data with the following results (p-values in parentheses):

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y^=0.044+0.006x1-0.035x2-0.001x3+0.296x4+0.010x5

(.070) (.228) (.157) (.678) (.001) (.869)

  1. Interpret the estimate of the β coefficient for x4.
  2. The “Big Bath” theory proposed by the researchers’ states that the mean DTVA for firms with negative earnings and earnings lower than last year will exceed the mean DTVA of other firms. Is there evidence to support this theory? Test using α = .05.
  3. Interpret the value of Ra2.

Drug content assessment. Refer to Exercise 8.16 (p. 467)and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatography to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different variances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values α=.05and to conduct the appropriate test for the researchers.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis α1α2=1

Alternative hypothesis α1α21

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

Question: Independent random samples n1 =233 and n2=312 are selected from two populations and used to test the hypothesis Ha:(μ1-μ)2=0against the alternative Ha:(μ1-μ)20

.a. The two-tailed p-value of the test is 0.1150 . Interpret this result.b. If the alternative hypothesis had been Ha:(μ1-μ)2<0 , how would the p-value change? Interpret the p-value for this one-tailed test.

Whistle-blowing among federal employees. Whistle blowing refers to an employee’s reporting of wrongdoing by co-workers. A survey found that about 5% of employees contacted had reported wrongdoing during the past 12 months. Assume that a sample of 25 employees in one agency are contacted and let x be the number who have observed and reported wrongdoing in the past 12 months. Assume that the probability of whistle-blowing is .05 for any federal employee over the past 12 months.

a. Find the mean and standard deviation of x. Can x be equal to its expected value? Explain.

b. Write the event that at least 5 of the employees are whistle-blowers in terms of x. Find the probability of the event.

c. If 5 of the 25 contacted have been whistle-blowers over the past 12 months, what would you conclude about the applicability of the 5% assumption to this

agency? Use your answer to part b to justify your conclusion.

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