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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
There is not enough evidence to conclude that loyalty to Chevrolet exceeds 47%.

Step by step solution

01

Define Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis, denoted as \( H_0 \), is that the proportion of Chevrolet owners loyal to the brand is equal to 0.47. The alternative hypothesis, \( H_a \), is that the proportion is greater than 0.47. Formally:\[H_0: p = 0.47 \quad \text{and} \quad H_a: p > 0.47\]
02

Compute Sample Proportion

The sample proportion \( \hat{p} \) is calculated as the number of loyal Chevrolet owners divided by the total sample size. Here, \( \hat{p} = \frac{490}{1006} \approx 0.487 \).
03

Calculate Standard Error

The standard error (SE) for the sample proportion is given by the formula: \[SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.47 \times (1 - 0.47)}{1006}} \approx 0.0157\]where \( n \) is the sample size and \( p \) is the claimed population proportion.
04

Compute Z-Score

The Z-score measures the number of standard errors the sample proportion is away from the hypothesized population proportion. It is calculated using:\[Z = \frac{\hat{p} - p}{SE} = \frac{0.487 - 0.47}{0.0157} \approx 1.084\]
05

Determine Critical Z-Value

For a right-tailed test at \( \alpha = 0.01 \), the critical Z-value is approximately 2.33 (found using Z-tables or statistical software).
06

Make a Decision

Compare the calculated Z-score with the critical Z-value. Since 1.084 is less than 2.33, we fail to reject the null hypothesis. This means there isn’t enough evidence to support the claim that the proportion of loyal Chevrolet consumers is more than 47% at the \(0.01\) significance level.

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Key Concepts

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

Understanding the Null Hypothesis
The null hypothesis is a fundamental component in hypothesis testing. It serves as a starting point for any statistical analysis. In most cases, the null hypothesis suggests that there is no effect or no difference in the population from what's already established. For the Chevrolet exercise, the null hypothesis ( \(H_0\) ) posits that the true proportion of brand-loyal Chevrolet consumers is exactly 47%, matching the general consumer population.

The null hypothesis assumes that any observed effect is due to random variation unless proven otherwise. In testing, it's the default position we seek to challenge through our sample data. This means if our sample doesn't provide strong enough evidence, we "fail to reject" it. It's crucial to note that not rejecting the null doesn’t confirm it as true, only that our evidence was insufficient to dismiss it.

Null hypotheses can be tested using various statistical tests, depending on the type of data and analysis required:
  • T-test: For comparing means.
  • Z-test: For comparing proportions or means in large samples, as seen in our scenario.
  • Chi-square tests: For testing relationships between categorical variables.
Exploring the Alternative Hypothesis
The alternative hypothesis offers a contrasting statement to the null hypothesis. It reflects what you aim to prove or find evidence for in your data. In the Chevrolet example, the alternative hypothesis ( \(H_a\) ) suggests that the true proportion of loyal Chevrolet consumers is greater than 47%.

The alternative hypothesis is the critical hypothesis under test. It is what researchers hope to support with data.
When formulating an alternative hypothesis, remember:
  • One-tailed vs. Two-tailed: Decide if you're looking for an effect in a specific direction (as in our one-tailed test where we're only interested in whether the proportion is greater, not just different) or in any direction (two-tailed).
  • Specificity: The hypothesis should be clear and precise in its statement to provide proper grounds for statistical testing.

In hypothesis testing, it's the alternative hypothesis you want to find enough evidence to "accept." If the test results suggest that the sample statistic is significantly different from the hypothesized population parameter (under \(H_0\) ), you reject the null in favor of the alternative.
The Role of Significance Level
The significance level, \( \alpha \) , in hypothesis testing determines the threshold for determining whether a result is statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true, commonly referred to as a Type I error. In the Chevrolet study, the significance level is set at 0.01, a stringent criterion.

Choosing a significance level involves a balance between being stringent enough to avoid false positives but flexible enough to detect true effects:
  • A typical value for \( \alpha \) is 0.05, but this can be adjusted depending on the context and the consequences of errors. In critical situations, such as medical trials, a lower \( \alpha \) like 0.01 is preferred to ensure higher accuracy.
  • The smaller the \( \alpha \) , the stronger the evidence required to reject the null hypothesis.

Using the significance level, one can find the critical value from statistical tables, such as a Z-table. If the calculated test statistic exceeds this critical value, the null hypothesis is rejected. In the exercise, our calculated Z-score (1.084) was compared against the critical value for \( \alpha = 0.01 \) . Because it did not exceed the critical value (2.33), we did not find sufficient evidence to support the claim that Chevrolet loyalty exceeds 47%.

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Most popular questions from this chapter

The body weight of a healthy 3 -month-old colt should be about \(\mu=60 \mathrm{~kg} .\) (Source: The Merck Veterinary Manual, a standard reference manual used in most veterinary colleges.) (a) If you want to set up a statistical test to challenge the claim that \(\mu=60 \mathrm{~kg}\), what would you use for the null hypothesis \(H_{0}\) ? (b) In Nevada, there are many herds of wild horses. Suppose you want to test the claim that the average weight of a wild Nevada colt \((3\) months old) is less than \(60 \mathrm{~kg} .\) What would you use for the alternate hypothesis \(H_{1} ?\) (c) Suppose you want to test the claim that the average weight of such a wild colt is greater than \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis? (d) Suppose you want to test the claim that the average weight of such a wild colt is different from \(60 \mathrm{~kg}\). What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the \(P\) -value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.

When conducting a test for the difference of means for two independent populations \(x_{1}\) and \(x_{2}\), what alternate hypothesis would indicate that the mean of the \(x_{2}\) population is smaller than that of the \(x_{1}\) population? Express the alternate hypothesis in two ways.

The pathogen Phytophthora capsici causes bell peppers to wilt and die. Because bell peppers are an important commercial crop, this disease has undergone a great deal of agricultural research. It is thought that too much water aids the spread of the pathogen. Two fields are under study. The first step in the research project is to compare the mean soil water content for the two fields (Source: Journal of Agricultural, Biological, and Environmental Statistics, Vol. 2, No. 2). Units are percent water by volume of soil. \(\begin{aligned} &\text { Field A samples, } x_{1} \text { : }\\\ &\begin{array}{llllrll} 10.2 & 10.7 & 15.5 & 10.4 & 9.9 & 10.0 & 16.6 \\ 15.1 & 15.2 & 13.8 & 14.1 & 11.4 & 11.5 & 11.0 \end{array} \end{aligned}\) \(\begin{aligned} &\text { Field B samples, } x_{2} \text { : }\\\ &\begin{array}{rrrrrrrr} 8.1 & 8.5 & 8.4 & 7.3 & 8.0 & 7.1 & 13.9 & 12.2 \\ 13.4 & 11.3 & 12.6 & 12.6 & 12.7 & 12.4 & 11.3 & 12.5 \end{array} \end{aligned}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx\) \(12.53, s_{1} \approx 2.39, \bar{x}_{2} \approx 10.77\), and \(s_{2} \approx 2.40\) ii. Assuming the distribution of soil water content in each field is moundshaped and symmetric, use a \(5 \%\) level of significance to test the claim that field \(\mathrm{A}\) has, on average, a higher soil water content than field \(\mathrm{B}\).

If we fail to reject (i.e., "accept") the null hypothesis, does this mean that we have proved it to be true beyond all doubt? Explain your answer.

USA Today reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents: \(\begin{array}{llllllllll} 72 & 68 & 81 & 93 & 56 & 19 & 78 & 94 & 83 & 84 \\ 77 & 69 & 85 & 97 & 75 & 71 & 86 & 47 & 66 & 27 \end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=\) \(71.4\) years and \(s\) \& \(20.65\) years. ii. Assuming that life span in Honolulu is approximately normally distributed, does this information indicate that the population mean life span for Honolulu residents is less than 77 years? Use a \(5 \%\) level of significance.

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