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Consumers: Product Loyalty USA Today reported that about \(47 \%\) of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose Chevrolet did a study of a random sample of 1006 Chevrolet owners and found that 490 said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal to Chevrolet is more than \(47 \% ?\) Use \(\alpha=0.01.\)

Short Answer

Expert verified
No, there's not enough evidence at \(\alpha = 0.01\) to conclude Chevrolet loyalty exceeds 47\%.

Step by step solution

01

Formulate Hypotheses

First, we need to set up our null and alternative hypotheses. The null hypothesis \( H_0 \) states that the population proportion \( p \) of Chevrolet owners loyal to the brand is equal to 0.47, the general population loyalty rate. Thus, \( H_0: p = 0.47 \). The alternative hypothesis \( H_a \) must reflect the claim that the population proportion is more than 0.47. Therefore, \( H_a: p > 0.47 \). We will use a significance level of \( \alpha = 0.01 \).
02

Gather Sample Data

From the study, we know that the sample size \( n = 1006 \), and the number of loyal Chevrolet owners \( X = 490 \). We can calculate the sample proportion \( \hat{p} = \frac{X}{n} = \frac{490}{1006} \approx 0.487 \).
03

Determine Standard Error

Calculate the standard error of the sample proportion using the formula: \( SE = \sqrt{\frac{p_0(1-p_0)}{n}} \), where \( p_0 = 0.47 \). So, \( SE = \sqrt{\frac{0.47 \times 0.53}{1006}} \approx 0.0157 \).
04

Compute Test Statistic

The test statistic \( z \) is calculated with the formula: \( z = \frac{\hat{p} - p_0}{SE} \). Substituting in our values: \( z = \frac{0.487 - 0.47}{0.0157} \approx 1.08 \).
05

Determine Critical Z-value

At \( \alpha = 0.01 \), for a one-tailed test, the critical z-value from standard normal tables is approximately 2.33. This is the value \( z_{critical} \) we need to exceed if the sample proportion significantly exceeds 0.47.
06

Make a Decision

Compare the test statistic \( z \approx 1.08 \) to the critical z-value of 2.33. Since \( 1.08 < 2.33 \), we fail to reject the null hypothesis.
07

Conclusion

We do not have enough evidence to support the claim that the proportion of Chevrolet owners who are loyal to the brand is greater than 47". At \( \alpha = 0.01 \), the test does not show a statistically significant increase in loyalty.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that suggests there is no effect or no difference in a given scenario. It's the position that any kind of observed effect is due to chance. The null hypothesis acts as the initial claim that you want to test. To illustrate this, in our Chevrolet loyalty example, the null hypothesis \( H_0 \) proposes that the actual proportion of consumer loyalty towards Chevrolet is equal to 47%, mirroring the general population's loyalty rate.

We write it mathematically as:
\( H_0: p = 0.47 \).

This establishes a baseline expectation. When conducting tests, especially in statistics, this is the hypothesis we attempt to disprove or reject. Thus, rejecting the null suggests that there is enough evidence to back up an alternative claim or hypothesis.

In this scenario, our aim is to analyze whether the observed loyalty rate truly exceeds this baseline of 47%.
Alternative Hypothesis
The alternative hypothesis is essentially the claim or theory that researchers seek to provide evidence for. It directly opposes the null hypothesis and represents a new perspective or expectation that one aims to demonstrate as true. In our Chevrolet loyalty study, the alternative hypothesis \( H_a \) posits that the proportion of Chevrolet loyalists exceeds the general 47%.

It can be written as:
\( H_a: p > 0.47 \).

This hypothesis suggests that more goodwill toward Chevrolet exists than previously known, indicating a stronger brand loyalty. The fact that the alternative hypothesis is generally the statement a researcher aims to substantiate makes it particularly relevant. If evidence supports the alternative hypothesis, this could have tangible consequences, influencing corporate strategies and market positioning.
  • The alternative hypothesis can be one-tailed, as in this case where we predict an increase, or two-tailed, where any deviation is tested.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by the researcher to determine if the findings from the statistical test are strong enough to reject the null hypothesis. This value essentially measures the probability of rejecting the null hypothesis when it is actually true.

In simpler terms, it reflects how much risk we are willing to take in terms of concluding a significant effect or difference where there might not be one. In our example, the significance level is set at 0.01, or 1%.

This choice of \( \alpha \) indicates we have very strict criteria for evidence, making it less likely to reject the null hypothesis unless the evidence is quite convincing. Choosing a lower significance level, like 0.01, means requiring more solid evidence before accepting the alternative hypothesis, while a higher level would allow for less stringent evidence leading to a potential conclusion.
Z-Test
The Z-test is a statistical method that helps determine whether two sample means are different when the variances are known and the sample size is large. It's particularly useful for testing hypotheses about population proportions when the sample size is large enough for the Central Limit Theorem to apply.

In the context of our Chevrolet loyalty analysis, a Z-test calculates how far, in terms of standard deviations, the sample proportion is from the hypothesized population proportion under the null hypothesis.

Here's how it works:
  • First, calculate the standard error, which serves as the standard deviation for the sampling distribution of the sample proportion.
  • Then, compute the Z-test statistic to see how many standard errors away the sample proportion is from the hypothesized 47%.

If this calculated Z-value is greater than the critical value from standard normal distribution tables, the null hypothesis may be rejected in favor of the alternative hypothesis. This process provides a basis for determining whether the observed data is significant.
Critical Value
Critical values are key elements in hypothesis testing. They act as benchmarks that define the cutoff points where the null hypothesis is rejected. These values correspond to the significance level and indicate the boundary for the extreme values within the null hypothesis distribution.

In the Chevrolet loyalty study, the critical Z-value at a 1% significance level for a one-tailed test is approximately 2.33.

This means that for any test statistic result beyond this critical value, there exists strong evidence against the null hypothesis, suggesting that the sample data is significantly inconsistent with the null assumption.
  • If the Z-test statistic calculated is less than 2.33, like in our example where it is 1.08, the conclusion is to fail to reject the null hypothesis.
  • If the result had been greater than 2.33, it might be reasonable to reject the null hypothesis and accept that there is more significant loyalty to Chevrolet than the general 47% population rate.

Critical values thus provide a crucial reference point in the decision-making process of hypothesis testing.

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Most popular questions from this chapter

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the \(z\) value of the sample test statistic. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in \(\mathrm{mg} / 100 \mathrm{ml}\) ). The sample mean is \(\bar{x} \approx 93.8 .\) Let \(x\) be a random variable representing glucose readings taken from Gentle Ben. We may assume that \(x\) has a normal distribution, and we know from past experience that \(\sigma=12.5 .\) The mean glucose level for horses should be \(\mu=85 \mathrm{mg} / 100 \mathrm{ml}\) (Reference: Merck Veterinary Manual). Do these data indicate that Gentle Ben has an overall average glucose level higher than \(85 ?\) Use \(\alpha=0.05\)

Suppose you want to test the claim that a population mean equals \(40 .\) (a) State the null hypothesis. (b) State the alternate hypothesis if you have no information regarding how the population mean might differ from \(40 .\) (c) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may exceed \(40 .\) (d) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may be less than \(40 .\)

Let \(x\) be a random variable that represents hemoglobin count (HC) in grams per 100 milliliters of whole blood. Then \(x\) has a distribution that is approximately normal, with population mean of about 14 for healthy adult women (see reference in Problem 17 ). Suppose that a female patient has taken 10 laboratory blood tests during the past year. The HC data sent to the patient's doctor are \(15 \quad 18$$ \quad$$16 \quad 19 \quad 14\) \(\begin{array}{ccccc}\quad12 & 14 & 17 & 15 & 11\end{array}\) i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}=15.1\) and \(s \approx 2.51.\) ii. Does this information indicate that the population average HC for this patient is higher than \(14 ?\) Use \(\alpha=0.01.\)

Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let \(c\) be the level of confidence used to construct a confidence interval from sample data. Let \(\alpha\) be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance \(\alpha\) and null hypothesis \(H_{0}\) : \(\mu=k,\) we reject \(H_{0}\) whenever \(k\) falls outside the \(c=1-\alpha\) confidence interval for \(\mu\) based on the sample data. When \(k\) falls within the \(c=1-\alpha\) confidence interval, we do not reject \(H_{0}\) (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as \(p, \mu_{1}-\mu_{2},\) and \(p_{1}-p_{2},\) which we will study in Sections 8.3 and \(8.5 .\) ) Whenever the value of \(k\) given in the null hypothesis falls outside the \(c=1-\alpha\) confidence interval for the parameter, we reject \(H_{0} .\) For example, consider a two-tailed hypothesis test with \(\alpha=0.01\) and $$H_{0}: \mu=20 \quad H_{1}: \mu \neq 20$$ A random sample of size 36 has a sample mean \(\bar{x}=22\) from a population with standard deviation \(\sigma=4.\) (a) What is the value of \(c=1-\alpha ?\) Using the methods of Chapter \(7,\) construct a \(1-\alpha\) confidence interval for \(\mu\) from the sample data. What is the value of \(\mu\) given in the null hypothesis (i.e., what is \(k\) )? Is this value in the confidence interval? Do we reject or fail to reject \(H_{0}\) based on this information? (b) Using methods of this chapter, find the \(P\) -value for the hypothesis test. Do we reject or fail to reject \(H_{0}\) ? Compare your result to that of part (a).

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Management: Lost Time In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable \(x_{1}\) measures a manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions: The variable \(x_{2}\) measures a manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable": i. Use a calculator with sample mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 4.86, s_{1} \approx 3.18, \bar{x}_{2}=6.5,\) and \(s_{2} \approx 2.88\) ii. Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use \(\alpha=0.05 .\) Assume that the two lost-time population distributions are mound-shaped and symmetric.

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