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Chapter 10: Problem 70

According to Pew Research Center surveys, \(79 \%\) of U.S. adults were using the Internet in January 2011 and \(83 \%\) were using it in January 2012 (USA TODAY, January 26,2012 ). Suppose that these percentages are based on random samples of 1800 U.S. adults in January 2011 and 1900 in January \(2012 .\) a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all U.S. adults who were using the Internet in January 2011 and January 2012, respectively. Construct a \(98 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is lower than \(p_{2}\) ? Use both the criticalvalue and the \(p\) -value approaches.

Short Answer

Expert verified
A. With 98% confidence interval, the difference between proportions \(p_{1} - p_{2}\) is calculated. B. With 1% significance level, decide whether \(p_{1}\) is less than \(p_{2}\) using critical value and p-value approaches for comparing proportions.

Step by step solution

01

Calculate \(p_1\) and \(p_2\)

The proportion \(p_1\) for January 2011 is calculated by multiplying 79 by 0.01, giving 0.79, likewise for January 2012, \(p_2\) is calculated by multiplying 83 by 0.01, giving 0.83.
02

Construct 98% Confidence Interval

The general formula for confidence interval is \((p_{1} - p_{2}) 卤 z * \sqrt{(p_{1}(1-p_{1})/n_1 + p_{2}(1-p_{2})/n_2)}\), where \(p_{1}, p_{2}\) are proportions, \(n_1, n_2\) are the sample sizes and \(z\) is the z-score for the desired confidence level. The z-score for a 98% confidence level is approximately 2.33. Substitute \(p_{1}=0.79, n_{1}=1800, p_{2}=0.83, n_{2}=1900\) into the formula to obtain the confidence interval.
03

Test Hypothesis using 1% significance level

The null hypothesis is \(p_{1} \geq p_{2}\). If the 98% confidence interval does not include zero, reject null hypothesis. If it does include zero, we fail to reject null hypothesis. To conclude, we use both critical value and p-value approaches.
04

Calculate p-value

The p-value can be calculated using a hypothesis test for comparing proportions.

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Key Concepts

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

Proportions
Proportions are an essential concept in statistics that refers to a type of ratio or fraction. When we talk about proportions in this context, we're looking at fractions of a total population or sample size. In our exercise, we have two proportions: one for Internet usage in January 2011 and another for January 2012.
  • The proportion for 2011 (\(p_1\)) is 0.79, meaning 79% of the sample was using the Internet.
  • For 2012 (\(p_2\)), it's 0.83, indicating an 83% Internet usage rate.
To estimate these proportions in the entire population, rather than just the sample, statistical methods like confidence intervals are used. Understanding this helps us draw conclusions about the larger group based on the sample data.
Hypothesis Testing
Hypothesis testing is a statistical method used for decision-making. It involves making an assumption, called a hypothesis, and then using sample data to determine whether there's enough evidence to support or reject the hypothesis. In our scenario, we are testing whether the proportion of people using the Internet (\(p_1\)) is lower in 2011 compared to 2012 (\(p_2\)). The null hypothesis (\(H_0\)) might state that there is no decrease, and \(p_1\) is greater than or equal to \(p_2\). On the other hand, the alternative hypothesis (\(H_a\)) suggests \(p_1\) is less than \(p_2\). The outcome of the hypothesis test involves checking whether it's reasonable to stick with the null hypothesis or to accept the alternative. This is where the significance level plays a crucial role.
Significance Level
The significance level, often denoted by \(\alpha\), is the threshold for deciding whether a result can be considered statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true. Common significance levels are 0.05 (5%) and 0.01 (1%). In this exercise, the significance level is 1%. This means we are looking for a very strong evidence to reject the null hypothesis. Choosing a lower significance level reduces the risk of a Type I error, which occurs when we wrongly reject a true null hypothesis. However, a lower \(\alpha\) makes it harder to prove the alternative hypothesis. It's a balance between detecting an effect and avoiding false positives.
P-value
The p-value is a calculated probability that helps in deciding on the null hypothesis. It indicates whether the observed data fit the null hypothesis. A low p-value suggests that the observed data would be very unlikely under the null hypothesis. To decide whether to reject \(H_0\), you compare the p-value against the significance level. If the p-value is less than or equal to \(\alpha\), you reject \(H_0\), indicating the data provide strong evidence against it. In our scenario, calculating the p-value helps us determine if the increase in Internet users from 2011 to 2012 is statistically significant. The lower the p-value, the stronger the evidence against \(H_0\). It's a key element in hypothesis tests, alongside the confidence intervals and critical values, to provide a comprehensive view of the hypothesis validity.

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Most popular questions from this chapter

Employees of a large corporation are concerned about the declining quality of medical services provided by their group health insurance. A random sample of 100 office visits by employees of this corporation to primary care physicians during 2004 found that the doctors spent an average of 19 minutes with each patient. This year a random sample of 108 such visits showed that doctors spent an average of \(15.5\) minutes with each patient. Assume that the standard deviations for the two populations are \(2.7\) and \(2.1\) minutes, respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means for these two years. b. Using a \(2.5 \%\) level of significance, can you conclude that the mean time spent by doctors with each patient is lower for this year than for \(2004 ?\) c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

According to a Randstad Global Work Monitor survey, \(52 \%\) of men and \(43 \%\) of women said that working part-time hinders their career opportunities (USA TODAY, October 6,2011 ). Suppose that these results are based on random samples of 1350 men and 1480 women. a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all men and all women, respectively, who will say that working part-time hinders their career opportunities. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(2 \%\) significance level, can you conclude that \(p_{1}\) and \(p_{2}\) are different? Use both the critical-value and the \(p\) -value approaches.

Assuming that the two populations have unequal and unknown population standard deviations, construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$

Perform the following tests of hypotheses, assuming that the populations of paired differences are normally distributed. a. \(H_{0} \cdot \mu_{d f}=0, \quad H_{1}: \mu_{d} \neq 0, \quad n=9, \quad \bar{d}=6.7, \quad s_{d}=2.5, \quad \alpha=.10\) b. \(H_{0}: \mu_{d}=0, \quad H_{1}: \mu_{d}>0, \quad n=22, \quad \bar{d}=14.8, \quad s_{d}=6.4, \quad \alpha=.05\) c. \(H_{0}=\mu_{d}=0, \quad H_{1}: \mu_{d}<0, \quad n=17, \quad \bar{d}=-9.3, \quad s_{d}=4.8, \quad \alpha=.01\)

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. Then the fuel consumption, in mpg, will be calculated for each test run. The average of the \(m p g\) for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg} .\) Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.c., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?
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