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Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The percentages of ontime arrivals for major U.S. airlines range from 68.6 to \(91.1 .\) Two regional airlines were surveyed with the following results. At \(\alpha=0.01\), is there a difference in proportions? $$ \begin{array}{lcc} & \text { Airline A } & \text { Airline B } \\ \hline \text { No. of flights } & 300 & 250 \\ \text { No. of on-time flights } & 213 & 185 \end{array} $$

Short Answer

Expert verified
No significant difference in on-time performance between Airline A and B at \(\alpha = 0.01\).

Step by step solution

01

State the Hypotheses and Identify the Claim

For this problem, we need to test if there is a difference in the on-time arrival rates between Airline A and Airline B. The null hypothesis \(H_0\) is that there is no difference in proportions: \(p_A = p_B\). The alternative hypothesis \(H_1\) is \(p_A eq p_B\). The claim here is that there is a difference in the on-time performance between the two airlines.
02

Find the Critical Value(s)

The level of significance \(\alpha = 0.01\) implies a two-tailed test to check if there is a difference in proportions. Using a standard normal distribution table or calculator for \(\alpha = 0.01\), the critical values are \(z = \pm 2.576\). This means we will reject the null hypothesis if the test statistic falls outside of the range between -2.576 and 2.576.
03

Compute the Test Value

First, calculate the sample proportions: \(\hat{p}_A = \frac{213}{300} = 0.71\) for Airline A and \(\hat{p}_B = \frac{185}{250} = 0.74\) for Airline B. Compute the pooled proportion: \(\hat{p} = \frac{213 + 185}{300 + 250} = \frac{398}{550} \approx 0.7236\). The standard error is calculated as: \[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = \sqrt{0.7236 \times (1 - 0.7236) \times \left(\frac{1}{300} + \frac{1}{250}\right)} \approx 0.0433.\] Then, calculate the test statistic:\[z = \frac{(\hat{p}_A - \hat{p}_B)}{SE} = \frac{(0.71 - 0.74)}{0.0433} \approx -0.692.\]
04

Make the Decision

The calculated z-value of approximately -0.692 does not fall in the critical region defined by \(z = \pm 2.576\). Therefore, we do not reject the null hypothesis \(H_0\).
05

Summarize the Results

Since we fail to reject the null hypothesis, we conclude that at the 0.01 significance level, there is not enough evidence to support the claim of a difference in the on-time performance proportions between Airline A and Airline B.

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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 key concept. It's essentially an initial assumption or statement that there is no effect or difference in the context of the problem being studied. For example, the null hypothesis

• Often symbolized as \( H_0 \) • Makes a baseline claim about a population parameter, like saying two groups are equal

In our example of comparing on-time flight rates between two airlines, the null hypothesis (\( H_0 \)) assumes both airlines have the same on-time rates, i.e., \( p_A = p_B \).
The main goal of hypothesis testing is to gather enough evidence to determine whether or not to reject this starting assumption. Depending on the data and test results, the null hypothesis can either be rejected in favor of an alternative hypothesis, or not rejected, suggesting there's insufficient evidence to make a new conclusion.
Critical Value
In hypothesis testing, a critical value helps in deciding whether to reject the null hypothesis. It's a threshold that the test statistic is compared against and serves as a marker dividing regions of acceptance from rejection.
Here's what you need to know:
    • The critical value is determined by the chosen significance level \( \alpha \), which indicates the probability of rejecting the null hypothesis when it's true • In a two-tailed test like ours, critical values occur at both extremes of the distribution
Given an \( \alpha = 0.01 \) significance level, we look up the critical values using a z-table for a two-tailed test. For our example, the critical values are \( z = \pm 2.576 \). This means if the calculated test statistic falls below \( -2.576 \) or above \( 2.576 \), we reject the null hypothesis and opt for the alternative one.
Standard Error
The standard error (SE) is an essential concept in statistics. It measures the variability or dispersion of a sampling distribution. In simpler terms, it tells us how much the sample statistic is expected to vary from the actual population parameter. The smaller the standard error, the closer the sample mean is to the true population mean.
Here's how to understand it:
    • Standard error is based on sample size and data variability • It's calculated differently depending on the data type and test
For our example, we needed the standard error of the difference in proportions between two airlines. The pooled proportion (\( \hat{p} \)) was calculated first and then used alongside sample sizes of Airline A and B to find SE:
\[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_A} + \frac{1}{n_B}\right)} \approx 0.0433.\]This SE value helped us determine the variability from which we calculated the test statistic, enabling a comparison against the critical value.
Test Statistic
The test statistic is a standardized value that measures the degree of deviation of the sample statistic from the null hypothesis. It’s a crucial part of hypothesis testing as it quantifies the evidence against the null hypothesis. The process involves comparing this statistic to a critical value to decide whether to reject the null hypothesis.
Here's a breakdown:
    • A higher or lower test statistic relative to the critical value can lead to rejecting the null • Test statistics vary by type of test (z, t, chi-square)
In our situation, we calculated a z-test statistic for the difference in flight on-time proportions:\[z = \frac{(\hat{p}_A - \hat{p}_B)}{SE} = \frac{(0.71 - 0.74)}{0.0433} \approx -0.692.\]This z-value represents how many standard errors the observed difference of sample proportions is away from the hypothesized difference of zero. In our case, \( -0.692 \) did not reach the critical region for \( z \), so we did not reject the null hypothesis.

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Most popular questions from this chapter

Adults aged 16 or older were assessed in three types of literacy: prose, document, and quantitative. The scores in document literacy were the same for 19 - to 24 -year-olds and for 40 - to 49 -year-olds. A random sample of scores from a later year showed the following statistics. $$ \begin{array}{lccc} & & \text { Population } & \\ \text { Age group } & \begin{array}{l} \text { Mean } \\ \text { score } \end{array} & \begin{array}{c} \text { standard } \\ \text { deviation } \end{array} & \begin{array}{c} \text { Sample } \\ \text { size } \end{array} \\ \hline 19-24 & 280 & 56.2 & 40 \\ 40-49 & 315 & 52.1 & 35 \end{array} $$ Construct a \(95 \%\) confidence interval for the true difference in mean scores for these two groups. What does your interval say about the claim that there is no difference in mean scores?

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. It has been found that \(26 \%\) of men 20 years and older suffer from hypertension (high blood pressure) and \(31.5 \%\) of women are hypertensive. A random sample of 150 of each gender was selected from recent hospital records, and the following results were obtained. Can you conclude that a higher percentage of women have high blood pressure? Use \(\alpha=0.05\) Men \(\quad 43\) patients had high blood pressure Women 52 patients had high blood pressure

The mean travel time to work for Americans is 25.3 minutes. An employment agency wanted to test the mean commuting times for college graduates and those with only some college. Thirty-five college graduates spent a mean time of 40.5 minutes commuting to work with a population variance of \(67.24 .\) Thirty workers who had completed some college had a mean commuting time of 34.8 minutes with a population variance of \(39.69 .\) At the 0.05 level of significance, can a difference in means be concluded?

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if the variance of the areas in square miles for counties in Indiana is less than the variance of the areas for counties in Iowa. A random sample of counties is selected, and the data are shown. At \(\alpha=0.01,\) can it be concluded that the variance of the areas for counties in Indiana is less than the variance of the areas for counties in Iowa? $$ \begin{array}{llll|llll} &&&{\text { Indiana }} &{\text { Iowa }} \\ \hline 406 & 393 & 396 & 485 & 640 & 580 & 431 & 416 \\ 431 & 430 & 369 & 408 & 443 & 569 & 779 & 381 \\ 305 & 215 & 489 & 293 & 717 & 568 & 714 & 731 \\ 373 & 148 & 306 & 509 & 571 & 577 & 503 & 501 \\ 560 & 384 & 320 & 407 & 568 & 434 & 615 & 402 \end{array} $$

The average sales price of new one-family houses in the Midwest is \(\$ 250,000\) and in the South is \(\$ 253,400\). A random sample of 40 houses in each region was examined with the following results. At the 0.05 level of significance, can it be concluded that the difference in mean sales price for the two regions is greater than \(\$ 3400 ?\) $$ \begin{array}{lll} & \text { South } & \text { Midwest } \\ \hline \text { Sample size } & 40 & 40 \\ \text { Sample mean } & \$ 261,500 & \$ 248,200 \\ \text { Population standard deviation } & \$ 10.500 & \$ 12.000 \end{array} $$

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