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Suppose that 1000 customers are surveyed and 850 are satisfied or very satisfied with a corporation's products and services. (a) Test the hypothesis \(H_{0}: p=0.9\) against \(H_{1}: p \neq 0.9\) at \(\alpha=0.05 .\) Find the \(P\) -value. (b) Explain how the question in part (a) could be answered by constructing a \(95 \%\) two-sided confidence interval for \(p\).

Short Answer

Expert verified
The hypothesis test shows the proportion is not 90%. A 95% CI (0.8314, 0.8686) supports this conclusion.

Step by step solution

01

Define the Hypotheses

To address part (a), we start by defining the null hypothesis and the alternative hypothesis. The null hypothesis is \( H_0: p = 0.9 \), which claims the proportion of satisfied customers is 90%. The alternative hypothesis is \( H_1: p eq 0.9 \), indicating that the true proportion differs from 90%.
02

Calculate the Test Statistic

First, find the sample proportion \( \hat{p} \). Calculate \( \hat{p} = \frac{number\ of\ successes}{sample\ size} = \frac{850}{1000} = 0.85 \). Then, calculate the standard error using \( SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.9 \times 0.1}{1000}} \approx 0.0095 \). The test statistic \( z \) is given by: \( z = \frac{\hat{p} - p}{SE} = \frac{0.85 - 0.9}{0.0095} \approx -5.263 \).
03

Determine the P-value

The computed test statistic is \( z = -5.263 \). Since this is a two-tailed test, we need the probability of observing a \( z \)-value as extreme or more extreme than -5.263. This P-value is found using standard normal distribution tables or software: \( P \text{-value} \approx 2 \times P(Z < -5.263) \), which is essentially 0, because -5.263 is far in the tails.
04

Make a Conclusion

Since the P-value (essentially 0) is less than \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \). There is significant evidence to suggest that the proportion of satisfied customers is not 90%.
05

Construct the 95% Confidence Interval for p

In part (b), to construct a 95% confidence interval for \( p \), use the formula: \( \hat{p} \pm Z_{\alpha/2} \cdot SE \). Here, \( Z_{\alpha/2} \approx 1.96 \) for a 95% confidence level. The interval is: \[ 0.85 \pm 1.96 \times 0.0095 \approx 0.85 \pm 0.0186 \]. The confidence interval is approximately (0.8314, 0.8686).
06

Compare Confidence Interval with Hypothesis Test

Since the 95% confidence interval for \( p \) (0.8314, 0.8686) does not include 0.9, this aligns with our hypothesis test result, indicating significant evidence that \( p eq 0.9 \).

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Key Concepts

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

Confidence Interval
When we talk about a confidence interval, we're really discussing a range of values that's likely to contain the true population parameter. In simpler terms, it's like saying, "Based on my data, I'm pretty sure the actual value falls within this range."
A confidence interval gives us both a lower and an upper limit. These limits are calculated from our sample data, and the level of confidence (like 95% in many cases) tells us how sure we are that the interval captures the true parameter.

For example, if you constructed a 95% confidence interval for a proportion, you'd expect that if you repeated the study many times, in 95% of the cases, the true proportion would fall within your interval. It's a powerful tool because it not only provides an estimate but also quantifies the uncertainty associated with that estimate.
In the exercise, the 95% confidence interval for the proportion of satisfied customers was found to be approximately (0.8314, 0.8686). This interval doesn't include 0.9, supporting the result of our hypothesis test that the actual satisfaction proportion isn't 90%.
Sample Proportion
The sample proportion is a straightforward concept: it represents the fraction of individuals in your sample that exhibit a particular characteristic. In our scenario, it's the proportion of satisfied customers among those surveyed.
The sample proportion, often denoted as \( \hat{p} \), is calculated by dividing the count of successes by the total sample size. So, if you have 850 satisfied customers out of 1000 surveyed, \( \hat{p} = \frac{850}{1000} = 0.85 \).

By studying the sample proportion, we can make inferences about the population proportion, which is the true proportion of all customers who are satisfied. Understanding the sample proportion is crucial before diving into more complex calculations, such as the test statistic or confidence interval, because it serves as the foundation for these analyses.
Standard Error
The standard error plays an essential role in statistics. It helps us understand how much our sample proportion might vary from the true population proportion. Think of it as a measure of the sample's "accuracy."
The standard error of a sample proportion is calculated with the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \]where \( p \) is the specified proportion under the null hypothesis, and \( n \) is the sample size. It tells you how much the sample proportion (\( \hat{p} \)) can be expected to vary from the true proportion just due to the randomness of sampling.

In the exercise, the standard error was about 0.0095. Recognizing this helps us to understand the variability of the sample proportion and is especially useful when constructing confidence intervals or calculating test statistics. The smaller the standard error, the more reliable the sample statistic is likely to be as an estimate of the population parameter.
Null Hypothesis
In hypothesis testing, the null hypothesis is essentially the status quo or a claim that there's no effect or no difference. It's the hypothesis we initially assume to be true until the data suggests otherwise.
The null hypothesis is denoted as \( H_0 \) and often includes a statement of equality, like \( p = 0.9 \). In the exercise, the null hypothesis is that the proportion of satisfied customers is exactly 90%.

The goal of hypothesis testing is to determine whether the data provides strong enough evidence to reject the null hypothesis. When we "reject \( H_0 \)," we conclude that the observed data is inconsistent with the null hypothesis, suggesting a real effect or difference exists.
Remember, rejecting the null doesn't prove the alternative hypothesis is true, but it does indicate that our assumption (the null) isn't likely correct. In our exercise, the null hypothesis \( H_0: p = 0.9 \) was rejected, showing significant evidence that the customer satisfaction rate isn't 90%.

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Most popular questions from this chapter

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