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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 propoctions 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 \(\alpha=. D 1\) and the following data: $$ \begin{aligned} &m=200 \quad \text { number of successes }-30 \\ &n=600 \text { number of successes }=180 \end{aligned} $$ (Similar dala is given in "Impact of Deals and Deal Retraction on Brand Switching" \(I\). Marketing, 19\$02: 62 - 70.)

Short Answer

Expert verified
Yes, financial inducement reduces loyalty more.

Step by step solution

01

Identify the Hypotheses

We are testing whether people induced to switch brands are less likely to remain loyal than those who switched without inducement. The null hypothesis is \(H_0: p_1 - p_2 = 0\), meaning there's no difference in loyalty. The alternative hypothesis is \(H_a: p_1 - p_2 < 0\), implying that those who switched with inducement are less loyal.
02

Determine Sample Proportions

Calculate the sample proportions of successes (repeat purchases) for each group. For those with inducement, \(\hat{p}_1 = \frac{30}{200} = 0.15\). For those without inducement, \(\hat{p}_2 = \frac{180}{600} = 0.3\).
03

Calculate the Test Statistic

Use the formula for the difference in proportions test statistic: \[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{m} + \frac{1}{n})}} \] where \(\hat{p} = \frac{x_1 + x_2}{m+n}\) with \(x_1 = 30\), \(x_2 = 180\), \(m = 200\), and \(n = 600\). Thus, \(\hat{p} = \frac{30+180}{800} = 0.2625\).
04

Calculate Pooled Sample Proportion and Standard Error

The pooled sample proportion \(\hat{p} = 0.2625\). Then, compute the standard error: \[ SE = \sqrt{0.2625(1-0.2625)(\frac{1}{200} + \frac{1}{600})} \approx 0.027615 \]
05

Find the Z-Statistic

Substitute the values into the formula for z: \[ z = \frac{0.15 - 0.3}{0.027615} \approx -5.427 \]
06

Decision Rule

At \(\alpha = 0.1\), we use the critical value method or the p-value approach. The z-critical value for a one-tailed test at \(\alpha = 0.1\) is approximately -1.28. Since \(z = -5.427\) is less than -1.28, we reject the null hypothesis.
07

Conclusion

Reject \(H_0\). There is significant evidence at \(\alpha = 0.1\) to suggest that those who switch brands due to financial inducement are less likely to remain loyal than those who switch without inducement.

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Key Concepts

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

Proportions Test
Proportions Test is a statistical method used to compare the proportion of a particular outcome in two different groups. It is especially useful when you want to determine if there is a significant difference in the proportion of successes between two groups. For example, in an exercise testing loyalty between those who switched brands with and without financial inducement, you would use a proportions test to see if there's a significant difference in loyalty.
To perform a proportions test, you will need:
  • The sample proportions for each group;
  • A hypothesis about the population proportions;
  • A confidence level, typically denoted by \( \alpha \).
The test involves calculating a "test statistic" that reflects how extreme the observed proportions are, under the assumption that the null hypothesis is true. If the test statistic exceeds a certain critical value, we may conclude there is enough evidence to support the alternative hypothesis.
Null Hypothesis
The Null Hypothesis, often represented as \( H_0 \), is a fundamental concept in hypothesis testing. It is the assumption that there is no effect or difference, meaning any observed difference is due to random sampling error. In the context of our exercise, the null hypothesis is stated as \( H_0: p_1 - p_2 = 0 \). This implies that there is no difference in loyalty between those who switched brands regardless of whether they received an inducement.
The null hypothesis is crucial because it provides a basis for comparison. By default, the assumption is that the null hypothesis is correct unless the evidence indicates otherwise. It minimizes the possibility of bias in drawing conclusions from the data. Hypothesis testing proceeds by either rejecting or failing to reject the null hypothesis based on the calculated test statistic.
Alternative Hypothesis
The Alternative Hypothesis, denoted as \( H_a \), suggests that there is an effect or a difference. Contrary to the null hypothesis, the alternative hypothesis indicates the presence of a statistically significant difference or effect in the data.
In our problem, the alternative hypothesis is expressed as \( H_a: p_1 - p_2 < 0 \). This means that those who switched brands due to financial inducement are hypothesized to be less loyal compared to those who switched without inducement. The nature of the alternative hypothesis dictates whether the test is one-tailed or two-tailed. Here, a one-tailed test is appropriate because the hypothesis is focused on a specific direction of effect.
Supporting the alternative hypothesis involves showing that our results are unlikely assuming the null hypothesis is true. Thus, if our test statistic is extreme enough, we have evidence to support the alternative.
Test Statistic
The Test Statistic is a crucial calculation in hypothesis testing. It is a numerical value that helps determine whether to reject the null hypothesis. The test statistic compares the observed data with what we would expect under the null hypothesis.
In a proportions test, the test statistic is calculated using the formula:\[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{m} + \frac{1}{n})}}\]Here, \( \hat{p} \) is the pooled sample proportion, derived from the combination of the sample observations from both groups. This formula essentially tells us how many standard errors away the difference in sample proportions is from zero. The further this value is from zero, the more evidence there is against the null hypothesis.After calculating this test statistic, it is compared against critical values to decide whether to reject the null hypothesis. If the computed \( z \) value is beyond the critical threshold, we have strong evidence that supports the alternative hypothesis.

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