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Redeeming tickets from textbook dispatches. Numerous companies now use textbook messaging on cell phones to sell their products. One way to do this is to shoot repairable reduction pasteboard (called an m- pasteboard) via a textbook. The redemption rate of m- tickets — the proportion of tickets redeemed — was the subject of a composition in the Journal of Marketing Research (October 2015). In a two-time study, over boardwalk shoppers shared by subscribing up to admit m-voucher. The experimenters were interested in comparing the redemption rates of m- tickets for different products in a sample of m- tickets for products vended at a milk-shake. Store, 79 were redeemed; in a sample of m- tickets for products vended at a donut store, 72 were redeemed.

a. Cipher the redemption rate for the sample of milk-shake m- tickets.

b. Cipher the redemption rate for the sample of donut m- tickets.

c. Give a point estimate for the difference between the actual redemption rates.

d. Form a 90 confidence interval for the difference between the actual redemption rates. Give a practical interpretation of the output.

e. Explain the meaning of the expression “90 confident” in your answer to part d.

f. Grounded on the interval, part d, is there a “ statistically.” A significant difference between the redemption rates? (Recall that a result is “ statistically” significant if there is substantiation to show that the true difference in proportions isn't 0.)

g. Assume the true difference between redemption rates must exceed.01 ( i.e., 1) for the experimenters to consider the difference “ virtually” significant. Based on the interval, part c, is there a “virtually” significant difference between the redemption rates?

Short Answer

Expert verified

The confidence interval (CI) is a set of values that are likely to encompass a population number with a given level of certainty.

Step by step solution

01

Step-by-Step Solution Step 1: (a) Find the redemption rate for the sample of the milk-shake m-voucher

Consider XM = 79 and nM=2,447.

The redemption proportion for the milk-shake m-voucher test is,

P¯M=xMnM

=792447

=0.032

Thus, the redemption rate for the sample of the milk-shake m-voucher is 0.032.

02

(b) Find the redemption rate for the sample donut m-voucher

Consider xD = 72 and nD = 6,619.

The redemption rate for the sample of donut m-voucher is,

PM=xDnD=726619

= 0.011

So, the reclamation rate for the sample of the donut m-voucher is 0.011.

03

(c) Find the point estimate for the difference between the actual redemption rates

PM-PD=0.032-0.011

= 0.021

Thus, the point estimate for the difference between the actual redemption rates is 0.021.

04

(d) Find the 90% confidence interval for the difference between the actual redemption rates

The critical value for a two-tailed test is obtained below:

Here, the test is two-tailed, and the significance level is α=0.10

The rejection region for the two-tailed test is|zc|>za2.

The confidence coefficient is 0.90.

So,

(1-α) = 0.90

α= 0.10

α2= 0.05

From Appendix D Table II, the critical value for the two-tailed test with α = 0.10 is α =0.10 is za2-0.05za2-0.05= ±1.645. so, the refusal region is |zc| > 1.645.

05

(e) 90% confidence interval

The 90% confidence interval is obtained below:

(P¯MP¯D)±z0.05P¯M(1P¯M)nM+P¯D(1P¯0)nD=0.021±1.6450.032(10.032)2447+0.011(10.011)6619

=0.021±1.645(0.00378)=0.021±0.006=(0.015,0.027)

Thus, the 90% confidence interval is (0.015, 0.027).

Interpretation:

There is 90% confidence that the difference between the actual redemption rates between milk-shake m-voucher and donut m-voucher lies between 0.015 and 0.027.

Explanation:

The 90% of all similarly generated confidence intervals will contain the true value of the difference in redemption rates in repeated sampling.

06

(f) check the statistically significant difference between the reclamation rates or not

Rule:

If the hypothesized value lies outside the corresponding 100(1-α) % confidence interval, then reject the null hypothesis.

Here, the 90% confidence interval is (0.015. 0.827), which does not contain the hypothesized value of 0. The hypothesized value 0 lies outside the interval (0.015, 0.027). Thus, it can be concluded that the null hypothesis H0 is rejected at = 0.10.

Hence, there is a statistically significant difference between the redemption rates.

07

(g) check the practically significant difference between the reclamation rates or not

If the genuine disparity among redeeming rates is more than 0.01, the disparity is considered "practically important."

Thus, there’s a "practically significant difference between the reclamation rates because the hypothesized value 0.01 lies outside the interval (0.015, 0.027).

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Most popular questions from this chapter

A paired difference experiment produced the following results:

nd=38,x¯1=92,x¯2=95.5,d¯=-3.5,sd2=21

a. Determine the values zfor which the null hypothesis μ1μ2=0would be rejected in favor of the alternative hypothesis μ1μ2<0 Use .role="math" localid="1652704322912" α=.10

b. Conduct the paired difference test described in part a. Draw the appropriate conclusions.

c. What assumptions are necessary so that the paired difference test will be valid?

d. Find a90% confidence interval for the mean difference μd.

e. Which of the two inferential procedures, the confidence interval of part d or the test of the hypothesis of part b, provides more information about the differences between the population means?

Find a value of the standard normal random variable z, call it z0, such that

a.P(zz0)=0.2090b.P(zz0)=0.7090c.P(-z0z<z0)=0.8472d.P(-z0z<z0)=0.1664e.P(z0zz0)=0.4798f.P(-1<z<z0)

Question: Consumers’ attitudes toward advertising. The two most common marketing tools used for product advertising are ads on television and ads in a print magazine. Consumers’ attitudes toward television and magazine advertising were investigated in the Journal of Advertising (Vol. 42, 2013). In one experiment, each in a sample of 159 college students were asked to rate both the television and the magazine marketing tool on a scale of 1 to 7 points according to whether the tool was a good example of advertising, a typical form of advertising, and a representative form of advertising. Summary statistics for these “typicality” scores are provided in the following table. One objective is to compare the mean ratings of TV and magazine advertisements.

a. The researchers analysed the data using a paired samples t-test. Explain why this is the most valid method of analysis. Give the null and alternative hypotheses for the test.

b. The researchers reported a paired t-value of 6.96 with an associated p-value of .001 and stated that the “mean difference between television and magazine advertising was statistically significant.” Explain what this means in the context of the hypothesis test.

c. To assess whether the result is “practically significant,” we require a confidence interval for the mean difference. Although this interval was not reported in the article, you can compute it using the information provided in the table. Find a 95% confidence interval for the mean difference and interpret the result. What is your opinion regarding whether the two means are “practically significant.”

Source: H. S. Jin and R. J. Lutz, “The Typicality and Accessibility of Consumer Attitudes Toward Television Advertising: Implications for the Measurement of Attitudes Toward Advertising in General,” Journal of Advertising, Vol. 42, No. 4, 2013 (from Table 1)

Is honey a cough remedy? Refer to the Archives of Pediatrics and Adolescent Medicine (December 2007) study of honey as a remedy for coughing, Exercise 2.31 (p. 86). Recall that the 105 ill children in the sample were randomly divided into groups. One group received a dosage of an over-the-counter cough medicine (DM); another group received a dosage of honey (H). The coughing improvement scores (as determined by the children’s parents) for the patients in the two groups are reproduced in the accompanying table. The pediatric researchers desire information on the variation in coughing improvement scores for each of the two groups.

a. Find a 90% confidence interval for the standard deviation in improvement scores for the honey dosage group.

b. Repeat part a for the DM dosage group.

c. Based on the results, parts a and b, what conclusions can the pediatric researchers draw about which group has the smaller variation in improvement scores? (We demonstrate a more statistically valid method for comparing variances in Chapter 8.)

Honey Dosage

11 12 15 11 10 13 10 13 10 4 15 16 9 14 10 6 10 11 12 12 8 12 9 11 15 10 15 9 13 8 12 10 9 5 12

DM Dosage

4 6 9 4 7 7 7 9 12 10 11 6 3 4 9 12 7 6 8 12 12 4 12 13 7 10 13 9 4 4 10 15 9

Fingerprint expertise.A study published in PsychologicalScience(August 2011) tested the accuracy of experts andnovices in identifying fingerprints. Participants were presentedpairs of fingerprints and asked to judge whetherthe prints in each pair matched. The pairs were presentedunder three different conditions: prints from the same individual (match condition), non-matching but similar prints (similar distracter condition), and nonmatching and very dissimilar prints (non-similar distracter condition). The percentages of correct decisions made by the two groups under each of the three conditions are listed in the table.

Conditions

Fingerprints expert

Novices

Match similar

92.12%

74.55%

Distracter

99.32%

44.82%

Non-similar distracter

100%

77.03%

a.Given a pair of matched prints, what is the probability that an expert failed to identify the match?

b. Given a pair of matched prints, what is the probabilitythat a novice failed to identify the match?

c. Assume the study included 10 participants, 5 experts and 5 novices. Suppose that a pair of matched prints was presented to a randomly selected study participant and the participant failed to identify the match. Is the participant more likely to be an expert or a novice?

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