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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

Drug content assessment. Scientists at GlaxoSmithKlineMedicines Research Center used high-performance liquidchromatography (HPLC) to determine the amountof drug in a tablet produced by the company (Analytical

Chemistry, Dec. 15, 2009). Drug concentrations (measuredas a percentage) for 50 randomly selected tablets are listedin the table below and saved in the accompanying file.

a. Descriptive statistics for the drug concentrations areshown at the top of the XLSTAT printout on the nextpage. Use this information to assess whether the dataare approximately normal.

b. An XLSTAT normal probability plot follows. Use thisinformation to assess whether the data are approximatelynormal.

91.28 92.83 89.35 91.90 82.85 94.83 89.83 89.00 84.62

86.96 88.32 91.17 83.86 89.74 92.24 92.59 84.21 89.36

90.96 92.85 89.39 89.82 89.91 92.16 88.67 89.35 86.51

89.04 91.82 93.02 88.32 88.76 89.26 90.36 87.16 91.74

86.12 92.10 83.33 87.61 88.20 92.78 86.35 93.84 91.20

93.44 86.77 83.77 93.19 81.79

Descriptive statistics(Quantitative data)

Statistic

Content

Nbr.of Observation

50

Minimum

81.79

Maximum

94.83

1st Quartile

87.2725

Median

89.375

3rd Quartile

91.88

Mean

89.2906

Variance(n-1)

10.1343

Standard deviation(n-1)

3.1834

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.

Given that xis a binomial random variable, compute P(x)for each of the following cases:

a. n= 7, x= 3, p= .5

b. n= 4, x= 3, p= .8

c. n= 15, x= 1, p= .1

Bankruptcy effect on U.S. airfares. Both Delta Airlines and USAir filed for bankruptcy. A study of the impact of bankruptcy on the fares charged by U.S. airlines was published in Research in Applied Economics (Vol. 2, 2010). The researchers collected data on Orlando-bound airfares for three airlines—Southwest (a stable airline), Delta (just entering bankruptcy at the time), and USAir (emerging from bankruptcy). A large sample of nonrefundable ticket prices was obtained for each airline following USAir’s emergence from bankruptcy, and then a 95% confidence interval for the true mean airfare was obtained for each. The results for 7-day advance bookings are shown in the accompanying table.

a. What confidence coefficient was used to generate the confidence intervals?

b. Give a practical interpretation of each of the 95% confidence intervals. Use the phrase “95% confident” in your answer.

c. When you say you are “95% confident,” what do you mean?

d. If you want to reduce the width of each confidence interval, should you use a smaller or larger confidence coefficient?

Assume that x is a binomial random variable with n = 1000 andp = 0.50. Use a normal approximation to find each of the following probabilities:

a. P(x>500)

b.P(490x<500)

c.P(x>550)

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