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91Ó°ÊÓ

Suppose that of 1000 customers surveyed, 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
Reject \(H_0\); P-value is very small. CI: \([0.8272, 0.8728]\), doesn't include 0.9.

Step by step solution

01

Understand the Hypotheses

In this hypothesis test, we are comparing a population proportion to a specific value. The null hypothesis is \(H_0: p = 0.9\), which suggests that 90% of the customers are satisfied. The alternative hypothesis \(H_1: p eq 0.9\) suggests that the actual proportion of satisfied customers is different from 90%.
02

Calculate Sample Proportion

The sample proportion \(\hat{p}\) is calculated by dividing the number of satisfied customers by the total number of surveyed customers. So, \(\hat{p} = \frac{850}{1000} = 0.85\).
03

Compute the Test Statistic

We'll use the test statistic for a proportion: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]where \(p_0 = 0.9\) and \(n = 1000\). Substituting the values, we have:\[Z = \frac{0.85 - 0.9}{\sqrt{\frac{0.9 \times 0.1}{1000}}} = \frac{-0.05}{0.00949} \approx -5.27\]
04

Determine the P-Value

Using the Z-table, we find the probability associated with \(Z = -5.27\). This Z-score is extremely low, indicating a very small P-value (much less than 0.05). This means there is strong evidence against the null hypothesis.
05

Compare P-Value to Significance Level

The P-value is significantly less than the significance level \(\alpha = 0.05\). Therefore, we reject the null hypothesis, \(H_0: p = 0.9\).
06

Construct a 95% Confidence Interval

To construct the confidence interval for \(p\):\[\hat{p} \pm Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\]where \(Z_{\alpha/2} = 1.96\) for a 95% CI. Substituting \(\hat{p}=0.85\) and \(n=1000\):\[0.85 \pm 1.96 \times \sqrt{\frac{0.85 \times 0.15}{1000}} = 0.85 \pm 0.0228 \]The confidence interval is \([0.8272, 0.8728]\).
07

Interpret the Confidence Interval

The interval \([0.8272, 0.8728]\) does not include 0.9, which suggests that 0.9 is not a plausible value for the true proportion based on the sample data, thus supporting the rejection of \(H_0\).

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Key Concepts

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

Population Proportion
Population proportion is a statistical measure that represents the fraction of the population exhibiting a certain characteristic. In our example, we are interested in the proportion of customers who are satisfied or very satisfied with a corporation's products and services. This is represented by the symbol \( p \). To find this proportion from a sample, you would divide the number of individuals in the sample exhibiting the characteristic by the total number of individuals in the sample. For example, if 850 out of 1000 customers surveyed are satisfied, the sample proportion (\( \hat{p} \)) is 0.85. Understanding population proportion is crucial because it provides insight into a population's characteristics based on a sample, which can be useful in decision-making. It serves as a foundational concept in hypothesis testing and helps to determine if the observed sample proportion is significantly different from a claimed or expected proportion.
Confidence Interval
A confidence interval provides a range of values that likely contain the true population proportion. This range gives us an idea of where the true proportion lies, with a specified level of confidence. For example, a 95% confidence interval means that we are 95% confident that the interval contains the true proportion.In the exercise, the confidence interval for the proportion of satisfied customers is calculated using the formula:\[\hat{p} \pm Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\]where \( \hat{p} \) is the sample proportion, and \( Z_{\alpha/2} \) is the Z-score corresponding to the desired confidence level. For a 95% confidence interval, \( Z_{\alpha/2} \) is typically 1.96.The confidence interval not only tells us about the possible values of the population proportion but also helps us see if a specific value (like 0.9 in our case) is a plausible population proportion based on the sample data.
Z-test for Proportion
The Z-test for proportion is a statistical test used to determine if there is a significant difference between the observed sample proportion and a specified population proportion. It's applicable when you are dealing with large sample sizes (n > 30 is a good rule of thumb). The test statistic is calculated as:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]where \( \hat{p} \) is the sample proportion, \( p_0 \) is the null hypothesis population proportion, and \( n \) is the sample size.In the example, we compared the sample proportion of 0.85 to the hypothesized proportion of 0.9. The Z-score of approximately -5.27 was calculated. A high absolute value of the Z-score indicates that the sample proportion differs significantly from the population proportion stated in the null hypothesis. Performing this test provides statistical evidence to either reject or fail to reject the null hypothesis.
P-value
The P-value in hypothesis testing measures the probability of observing test results at least as extreme as the results that were actually observed, under the assumption that the null hypothesis is true. A smaller P-value indicates stronger evidence against the null hypothesis. In most educational contexts, a P-value less than 0.05 is considered significant. This means that there's a low probability (less than 5%) that the observed sample proportion would occur if the null hypothesis were true. In our exercise, the P-value is significantly smaller than the significance level of 0.05, which leads us to reject the null hypothesis. This conclusion is drawn because the observed sample proportion provides strong evidence that the true population proportion differs from the hypothesized value of 0.9.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set before conducting a hypothesis test. It's the probability of rejecting the null hypothesis when it is, in fact, true (a type I error). Commonly used significance levels are 0.05, 0.01, and 0.10.In our example, the significance level is set at 0.05. This means that you are willing to accept a 5% risk of concluding that the population proportion is different from the hypothesized proportion when it is not.The significance level helps to decide whether the evidence from the sample data is strong enough to reject the null hypothesis. If the P-value calculated from the test is less than the \( \alpha \), it suggests that the null hypothesis should be rejected because there is sufficient evidence to support the alternative hypothesis.

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Most popular questions from this chapter

A biotechnology company produces a therapeutic drug whose concentration has a standard deviation of 4 grams per liter. A new method of producing this drug has been proposed, although some additional cost is involved. Management will authorize a change in production technique only if the standard deviation of the concentration in the new process is less than 4 grams per liter. The researchers chose \(n=10\) and obtained the following data in grams per liter. Perform the necessary analysis to determine whether a change in production technique should be implemented. $$\begin{array}{ll}16.628 & 16.630 \\\16.622 & 16.631 \\\16.627 & 16.624 \\\16.623 & 16.622 \\\16.618 & 16.626\end{array}$$

Output from a software package follows: Test of \(m u=99\) vs \(>99\) The assumed standard deviation \(=2.5\) $$\begin{array}{ccccccc}\text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\\x & 12 & 100.039 & 2.365 & ? & 1.44 & 0.075 \\\\\hline\end{array}$$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) If the hypothesis had been \(H_{0}: \mu=98\) versus \(H_{0}: \mu>98\), would you reject the null hypothesis at the 0.05 level of significance? Can you answer this without referring to the normal table? (d) Use the normal table and the preceding data to construct a \(95 \%\) lower bound on the mean. (e) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 99 ?\)

State the null and alternative hypothesis in each case. (a) A hypothesis test will be used to potentially provide evidence that the population mean is more than \(10 .\) (b) A hypothesis test will be used to potentially provide evidence that the population mean is not equal to 7 . (c) A hypothesis test will be used to potentially provide evidence that the population mean is less than \(5 .\)

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20\), approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)

A quality-control inspector is testing a batch of printed circuit boards to see whether they are capable of performing in a high temperature environment. He knows that the boards that will survive will pass all five of the tests with probability \(98 \% .\) They will pass at least four tests with probability \(99 \%,\) and they always pass at least three. On the other hand, the boards that will not survive sometimes pass the tests as well. In fact, \(3 \%\) pass all five tests, and another \(20 \%\) pass exactly four. The rest pass at most three tests. The inspector decides that if a board passes all five tests, he will classify it as "good." Otherwise, he'll classify it as "bad." (a) What does a type I error mean in this context? (b) What is the probability of a type I error? (c) What does a type II error mean here? (d) What is the probability of a type II error?

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