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Chapter 10: Problem 48

Is someone who switches brands because of a financial inducement less likely to remain loyal than someone who switches without inducement? Let \(p_{1}\) and \(p_{2}\) denote the true proportions of switchers to a certain brand with and without inducement, respectively, who subsequently make a repeat purchase. Test \(H_{0}: p_{1}-p_{2}=0\) versus \(H_{a}: p_{1}-p_{2}<0\) using \(\alpha=.01\) and the following data: $$ \begin{array}{ll} m=200 & \text { number of successes }=30 \\ n=600 & \text { number of successes }=180 \end{array} $$ (Similar data is given in "Impact of Deals and Deal Retraction on Brand Switching," J. Marketing. \(1980: 62-70 .\) )

Short Answer

Expert verified
The person switching with inducement is less likely to remain loyal.

Step by step solution

01

Identify Given Data

We are given that the sample size for the group with inducement, \(m\), is 200 with 30 successes. For the group without inducement, \(n\), is 600 with 180 successes. We need to find whether there is a significant difference in proportions.
02

Calculate Sample Proportions

Calculate the sample proportions \( \hat{p}_1 \) and \( \hat{p}_2 \) for each group. \( \hat{p}_1 = \frac{30}{200} = 0.15 \) and \( \hat{p}_2 = \frac{180}{600} = 0.3 \).
03

Formulate Hypotheses

The null hypothesis is \( H_0: p_1 - p_2 = 0 \) and the alternative hypothesis is \( H_a: p_1 - p_2 < 0 \). We are testing if the proportion with inducement is less than without inducement.
04

Calculate Pooled Proportion

The pooled proportion \( \hat{p} \) is calculated as \( \hat{p} = \frac{x_1 + x_2}{m + n} = \frac{30 + 180}{200 + 600} = \frac{210}{800} = 0.2625 \).
05

Calculate Standard Error

Calculate the standard error (SE) using the formula \( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{m} + \frac{1}{n}\right)} = \sqrt{0.2625(1-0.2625)\left(\frac{1}{200} + \frac{1}{600}\right)} \approx 0.034 \).
06

Compute Test Statistic

Find the test statistic \( z \) using \( z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.15 - 0.3}{0.034} \approx -4.412 \).
07

Determine Critical Value

Since \( \alpha = 0.01 \) and this is a one-tailed test, find the critical value for \( z \) from the standard normal distribution, which is approximately \( -2.33 \).
08

Compare and Conclude

Compare \( z \) with the critical value: \(-4.412 < -2.33\). Since \( z \) is less than the critical value, we reject the null hypothesis in favor of the alternative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Test
A proportion test is a statistical tool used to determine if there is a significant difference between two proportions. In the given exercise, we want to find out if people who switched brands due to financial inducements have a different likelihood of making a repeat purchase compared to those who switched without inducements.

To carry out the test, we start by estimating the sample proportions of both groups. These proportions are calculated from the number of successes and the sample size for each group. The sample proportions are then compared to assess if there's a noticeable difference between the two populations represented in the sample.
  • The null hypothesis (\( H_0 \)) suggests there's no difference in proportions.
  • The alternative hypothesis (\( H_a \)) proposes a less than relationship, indicating the proportion with inducement is smaller than without.

Setting up these hypotheses is crucial, as they guide the analysis and affect the interpretation of results.
Statistical Significance
Statistical significance helps us understand whether the observed difference between two proportions is due to random chance or if it reflects a true difference in the population.

In hypothesis testing, we use statistical significance to decide whether to accept or reject the null hypothesis. This involves comparing a calculated test statistic to a critical value, which depends on our chosen significance level (\( \alpha \)). In our exercise, \( \alpha \) is set at 0.01, meaning we are allowing a 1% probability of rejecting the null hypothesis incorrectly.
  • A test statistic further from zero indicates a more significant difference between proportions.
  • If the test statistic is beyond the critical value, we reject the null hypothesis.
  • This result tells us the observed difference is unlikely due to chance, confirming a significant difference in loyalty between the two groups.

Understanding statistical significance is vital to interpreting the outcome of a hypothesis test correctly.
Pooled Proportion
The pooled proportion is a combined estimate of the proportion when two samples are considered as a single group. It's a weighted average of the sample proportions, which provides a best estimate of the population proportion under the null hypothesis where no difference exists.

This concept is used primarily to calculate the standard error in a two-proportion test. The pooled proportion is calculated using the total number of successes divided by the total sample size from both groups combined.
  • This approach assumes that under the null hypothesis, the true proportions are essentially the same.
  • It provides a baseline from which to measure the variance of the observed sample proportions.

Utilizing the pooled proportion is a crucial step when performing two-sample proportion tests as it impacts the calculation of variability, which feeds into the test statistic.
Standard Error
In hypothesis testing, specifically for proportions, the standard error measures how much the sample proportion is likely to vary from the population proportion.

The standard error is calculated using the pooled proportion, which is why it's important to estimate this value accurately. It accounts for the size of both samples and their outcomes, providing a measure of the precision of our sample estimate.
  • The formula for standard error in two-sample proportion tests is: \[ SE = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{m} + \frac{1}{n})} \]
  • A smaller standard error indicates more reliable and stable estimates of the population proportion.

By understanding the standard error, we can make informed judgments about the uncertainty of our results, helping us interpret whether observed differences in sample proportions are statistically significant.

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Most popular questions from this chapter

A student project by Heather Kral studied students on "lifestyle floors" of a dormitory in comparison to students on other floors. On a lifestyle floor the students share a common major, and there are a faculty coordinator and resident assistant from that department. Here are the grade point averages of 30 students on lifestyle floors (L) and 30 students on other floors \((\mathrm{N})\) : L: \(2.00,2.25,2.60,2.90,3.00,3.00,3.00,3.00\), \(3.00,3.20,3.20,3.25,3.30,3.30,3.32,3.50\), \(3.50,3.60,3.60,3.70,3.75,3.75,3.79,3.80\), \(3.80,3.90,4.00,4.00,4.00,4.00\). \(\mathrm{N}: 1.20,2.00,2.29,2.45,2.50,2.50,2.50,2.50\), \(2.65,2.70,2.75,2.75,2.79,2.80,2.80,2.80\), \(2.86,2.90,3.00,3.07,3.10,3.25,3.50,3.54\), \(3.56,3.60,3.70,3.75,3.80,4.00\). Notice that the lifestyle grade point averages have a large number of repeats and the distribution is skewed, so there is some question about normality. a. Obtain a \(95 \%\) confidence interval for the difference of population means using the method based on the theorem of Section 10.2. b. Obtain a bootstrap sample of 999 differences of means. Check the bootstrap distribution for normality using a normal probability plot. c. Use the standard deviation of the bootstrap distribution along with the mean and \(t\) critical value from (a) to get a \(95 \%\) confidence interval for the difference of means. d. Use the bootstrap sample and the percentile method to obtain a \(95 \%\) confidence interval for the difference of means. e. Compare your three confidence intervals. If they are very similar, why do you think this is the case? f. Interpret your results. Is there a substantial difference between lifestyle and other floors? Why do you think the difference is as big as it is?

The accompanying data on the alcohol content of wine is representative of that reported in a study in which wines from the years 1999 and 2000 were randomly selected and the actual content was determined by laboratory analysis (London Times, Aug. 5, 2001). $$ \begin{array}{lcccccc} \text { Wine } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Actual } & 14.2 & 14.5 & 14.0 & 14.9 & 13.6 & 12.6 \\ \text { Label } & 14.0 & 14.0 & 13.5 & 15.0 & 13.0 & 12.5 \end{array} $$ The two-sample \(t\) test gives a test statistic value of \(.62\) and a two-tailed \(P\)-value of \(.55\). Does this convince you that there is no significant difference between true average actual alcohol content and true average content stated on the label? Explain.

Sometimes experiments involving success or failure responses are run in a paired or before/ after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: where \(X_{1}+X_{2}+X_{3}+X_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(P_{4}\) denote the four cell probabilities, so that \(P_{1}=P(S\) before and \(S\) after \()\), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{i}-X_{j}\right) / n\) has approximately a normal distribution with variance \(\left[p_{i}+p_{j}-\left(p_{i}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called MeNemar's test). d. If \(x_{1}=350, x_{2}=150, x_{3}=200\), and \(x_{4}=300\), what do you conclude?

The level of monoamine oxidase (MAO) activity in blood platelets \((\mathrm{nm} / \mathrm{mg}\) protein/ \(\mathrm{h})\) was determined for each individual in a sample of 43 chronic schizophrenics, resulting in \(\bar{x}=2.69\) and \(s_{1}=2.30\), as well as for 45 normal subjects, resulting in \(\bar{y}=6.35\) and \(s_{2}=4.03\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha=.01\). [Hint: \(H_{0}\) and \(H_{\mathrm{a}}\) here have a different form from the three standard cases. Let \(\mu_{1}\) and \(\mu_{2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta=2 \mu_{1}-\mu_{2}\). Write \(H_{0}\) and \(H_{\mathrm{a}}\) in terms of \(\theta\), estimate \(\theta\), and derive \(\hat{\sigma}_{\hat{\theta}}\) ("Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients," Nature, July 28, 1972: 225-226).]

An experiment was carried out to compare various properties of cotton/polyester spun yarn finished with softener only and yarn finished with softener plus 5\% DP-resin ("Properties of a Fabric Made with Tandem Spun Yarns," Textile Res. \(J ., 1996: 607-611)\). One particularly important characteristic of fabric is its durability, that is, its ability to resist wear. For a sample of 40 softener-only specimens, the sample mean stoll-flex abrasion resistance (cycles) in the filling direction of the yarn was \(3975.0\), with a sample standard deviation of \(245.1\). Another sample of 40 softener-plus specimens gave a sample mean and sample standard deviation of \(2795.0\) and \(293.7\), respectively. Calculate a confidence interval with confidence level \(99 \%\) for the difference between true average abrasion resistances for the two types of fabrics. Does your interval provide convincing evidence that true average resistances differ for the two types of fabrics? Why or why not?
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