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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.

Short Answer

Expert verified
a) The 95% confidence interval for the difference between these two population means is computed in Step 1. b) If the calculated Z score is less than -1.96, we conclude at the 2.5% level of significance that the mean time spent by doctors this year is significantly lower than in 2004. c) If the probability of a Type I error is zero, we do not reject the null hypothesis and therefore, we cannot conclude that the mean time spent by doctors this year is significantly lower than in 2004.

Step by step solution

01

Construct the 95% confidence interval

The first task here is to construct a 95% confidence interval for the difference between the two population means. For this, we utilize the formula for confidence interval of difference between two means: \[CI = (\bar{x_1} - \bar{x_2}) \pm (Z \cdot \sqrt{(\sigma_1^2/n_1) + (\sigma_2^2/n_2)})\], where \(\bar{x_1}\) and \(\bar{x_2}\) are sample means, \(n_1\) and \(n_2\) are sample sizes, \(\sigma_1\) and \(\sigma_2\) are standard deviations, and Z is the Z value from the standard normal distribution for the required confidence level. For a 95% confidence level, Z is approximately 1.96. Substituting all the given values, we get \[CI = (19 - 15.5) \pm (1.96 \cdot \sqrt{(2.7^2/100) + (2.1^2/108)})\]
02

Test the hypothesis at the 2.5% level of significance

Next, we use the given level of significance to test the hypothesis that this year's mean time is less than that of 2004. For this, we calculate the Z score as follows: \[Z = (\bar{x_1} - \bar{x_2}) / \sqrt{(\sigma_1^2/n_1) + (\sigma_2^2/n_2)}\] Substitute the given values to find the Z score. If the Z score is less than -1.96 (the critical Z value for a one-sided test at the 2.5% level of significance), we reject the null hypothesis that the means are equal and conclude that the mean time spent by doctors with patients this year is significantly lower than in 2004.
03

Decision if probability of Type I error is zero

If the probability of a Type I error is zero, then the decision would be to not reject the null hypothesis, irrespective of the value of the Z score. The rationale behind this decision is based on the fact that a Type I error (false positive) is considered a more serious mistake and we simply cannot afford to commit such an error. This means sticking to the null hypothesis and acting conservatively upon our analysis. Thus, we cannot conclude that the mean time spent by doctors this year is significantly lower than in 2004.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics used to make inferences about populations based on sample data. It involves making a claim, known as the null hypothesis ( ), and testing whether this claim can be rejected in favor of an alternative hypothesis. An important aspect of hypothesis testing is choosing the appropriate test based on the data and context. In the original exercise, the null hypothesis is that the mean time doctors spend with patients this year is the same as in 2004. The alternative hypothesis is that the time is lower this year. By calculating a test statistic, such as a Z score, and comparing it to a critical value set by our level of significance, we decide whether to reject the null hypothesis. A lower calculated Z score than the critical value indicates that we reject the null hypothesis and accept the alternative. Hypothesis testing helps in making informed decisions from sample data by providing a structured framework to assess evidence.
Standard Deviation
Standard deviation is a measure of the amount of variation in a set of data values. It tells us how much the numbers in a data set deviate from the mean of the data set. A smaller standard deviation indicates that the data points are close to the mean, whereas a larger standard deviation indicates that they are spread out over a wider range of values. In this exercise, the standard deviations are used to calculate the confidence interval and conduct hypothesis testing. They help determine the spread of individual doctor visit times around their average for each year. This spread is crucial for accurately estimating how much the population means might differ between the two years. For instance, in calculating a 95% confidence interval or a Z score for hypothesis testing, the standard deviations are used to understand the reliability and variability in our samples. They ensure our conclusions about population differences are justified.
Type I Error
A Type I error occurs when a true null hypothesis is incorrectly rejected. This is also known as a "false positive." It means that based on the sample data, we believe there is an effect or difference when in fact there isn't one. In the context of hypothesis testing in the exercise, a Type I error would mean concluding that doctors are spending less time with patients this year compared to 2004, even though this is not true. The significance level ( alpha) sets the probability of making a Type I error. A lower alpha value reduces the chance of committing this error, but also increases the likelihood of a Type II error (failing to reject a false null hypothesis). The original exercise highlights that if the probability of a Type I error is zero, we would not reject the null hypothesis regardless of the evidence. This scenario underscored the care taken in statistical hypothesis testing to avoid false positives.
Population Means
Population means represent the average of a characteristic in a population. In statistics, when direct measurement of all members of a population is not feasible, sample means are used to estimate and make inferences about the population means. The exercise involves comparing the population means for doctor visit duration in two different years. Since recording the exact mean time for every patient visit is impractical, samples were used to estimate the means for 2004 and this year. The confidence interval provides a range of values within which we expect the true difference between the population means to lie, with a certain level of confidence (e.g., 95%). Accurate estimation of population means through samples allows organizations to understand broader trends and changes, such as assessing if the time spent with doctors has statistically decreased this year compared to 2004.

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Most popular questions from this chapter

Construct a 99 ?o confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=300, \quad \hat{p}_{1}=.55, \quad n_{2}=200, \quad \hat{p}_{2}=.62 $$

Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=100, \quad \hat{p}_{1}=.81, \quad n_{2}=150, \quad \hat{p}_{2}=.77 $$

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In parts of the eastern United States, whitctail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent \(\mathrm{A}\), and the other 27 receive repellent \(\mathrm{B}\). For each of the 56 gardens, the time elapsed between application of the repellent and the appearance in the garden of the first deer is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(B\), the mean time is 92 hours. Assume that the two populations of elapsed times have normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) c. Test at a \(2 \%\) significance level whether the mean elapsed times for repellents \(A\) and \(B\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.

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