/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Researchers have shown that ciga... [FREE SOLUTION] | 91Ó°ÊÓ

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Researchers have shown that cigarette smoking has a deleterious effect on lung function. In their study of the effect of cigarette smoking on the carbon monoxide diffusing capacity (DL) of the lung, Ronald Knudson, W. Kaltenborn and B. Burrows found that current smokers had DL readings significantly lower than either ex-smokers or nonsmokers. \(^{*}\) The carbon monoxide diffusing capacity for a random sample of current smokers was as follows: $$\begin{array}{rrrrr} 103.768 & 88.602 & 73.003 & 123.086 & 91.052 \\ 92.295 & 61.675 & 90.677 & 84.023 & 76.014 \\ 100.615 & 88.017 & 71.210 & 82.115 & 89.222 \\ 102.754 & 108.579 & 73.154 & 106.755 & 90.479 \end{array}$$ Do these data indicate that the mean DL reading for current smokers is lower than 100 , the average DL reading for nonsmokers? a. Test at the \(\alpha=.01\) level. b. Bound the \(p\) -value using a table in the appendix. c. Find the exact \(p\) -value.

Short Answer

Expert verified
Reject the null hypothesis; the mean DL for smokers is less than 100 (p < 0.01).

Step by step solution

01

Calculate Sample Mean

First, we need to calculate the mean DL reading for the sample of current smokers. Add up all the DL readings and divide by the total number of readings. The readings given are:\[103.768, 88.602, 73.003, 123.086, 91.052, 92.295, 61.675, 90.677, 84.023, 76.014, 100.615, 88.017, 71.210, 82.115, 89.222, 102.754, 108.579, 73.154, 106.755, 90.479\]The sum of these values is 1725.08, and there are 20 readings in total.\[\bar{x} = \frac{1725.08}{20} = 86.254\]
02

Calculate Sample Standard Deviation

Next, we calculate the standard deviation of the sample. The formula for sample standard deviation is:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\]Calculate each deviation from the mean, square it, sum all squared deviations, divide by \(n-1 = 19\), and take the square root.After calculations, we find:\[s = 15.709\]
03

Determine Test Statistic

We use a one-sample t-test for the mean to determine if the mean DL reading is significantly lower than 100. The test statistic is computed as:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]Where \(\mu = 100\), \(\bar{x} = 86.254\), \(s = 15.709\), and \(n = 20\).\[t = \frac{86.254 - 100}{15.709 / \sqrt{20}} = -3.505\]
04

Compare with Critical Value and Determine Result

The critical value for a one-tailed t-test at \(\alpha = 0.01\) with \(n-1 = 19\) degrees of freedom can be found in a t-distribution table. It is approximately \(-2.539\). Since \(-3.505 < -2.539\), we reject the null hypothesis.
05

Bound the p-value

Using a t-distribution table, the p-value for \(t = -3.505\) with 19 degrees of freedom is less than the p-value associated with the t-score limit for \(\alpha = 0.01\) (critical value = -2.539), confirming it's less than 0.01.
06

Calculate Exact p-value

We use statistical software or an online calculator for an exact p-value calculation for the t-distribution with 19 degrees of freedom: The exact p-value is approximately 0.0012.

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Key Concepts

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

t-distribution
The t-distribution is crucial when dealing with small sample sizes, typically 30 or fewer observations. Unlike the normal distribution, the t-distribution has heavier tails, which means it is more probable to obtain values far away from the mean. This property makes it suitable for estimating population parameters when the sample size is small, and the population standard deviation is unknown.

In hypothesis testing, the t-distribution helps determine how far the sample mean deviates from the hypothesized population mean. The shape of the t-distribution depends on the degrees of freedom (df), which is calculated as the sample size minus one ( -1) when assessing a single group. As the degrees of freedom increase, the t-distribution gets closer to the normal distribution.

In the context of the exercise, the t-distribution plays a pivotal role in deciding whether the mean carbon monoxide diffusing capacity of current smokers is significantly different from nonsmokers. A t-value is computed and compared against a critical value from the t-distribution table to make a decision regarding the null hypothesis.
p-value
The p-value is a significant concept in statistical hypothesis testing. It represents the probability of observing the given data, or something more extreme, if the null hypothesis is true. In simpler terms, it helps in determining whether the evidence is strong enough to reject the null hypothesis.

A lower p-value indicates stronger evidence against the null hypothesis. Typically, we compare the p-value to a pre-specified significance level (such as 0.01, 0.05) to decide whether to reject or fail to reject the null hypothesis. If the p-value is less than the significance level, we generally reject the null hypothesis.

In the exercise, a p-value is calculated to evaluate whether the mean DL reading for current smokers is significantly lower than that of nonsmokers. The exact p-value obtained significantly less than 0.01 suggests that smoking indeed affects the carbon monoxide diffusing capacity, corroborating the researchers' findings.
sample standard deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of values. It provides insight into how spread out the sample data is around the sample mean. The formula for calculating the sample standard deviation is given by \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\]where \(x_i\) denotes each data point, \(\bar{x}\) is the sample mean, and \(n\) is the number of observations in the sample.

Calculating the sample standard deviation involves finding the difference between each data point and the mean, squaring these differences, summing them, and then dividing by \(n - 1\). Taking the square root of this result yields the sample standard deviation.

In the given study, the calculated sample standard deviation is used in the one-sample t-test to understand the variability within the lung function readings of the smokers. It is important because it influences the value of the t-statistic, helping to determine whether the observed sample mean significantly deviates from the population mean.
one-sample t-test
The one-sample t-test is a statistical method used to determine whether the mean of a single sample significantly differs from a known or hypothetical population mean. This test is appropriate when dealing with small sample sizes from a normally-distributed population where the population standard deviation is not known.

To perform a one-sample t-test, compute the t-statistic using:\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

In the exercise, the researchers carried out a one-sample t-test to ascertain whether the average DL reading for current smokers was significantly less than 100. The calculated t-statistic was then compared with critical values from the t-distribution. A significant result indicated a lower mean DL reading for smokers compared to the average nonsmoker reading, suggesting negative effects of smoking on lung function.

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Most popular questions from this chapter

An Article in American Demographics investigated consumer habits at the mall. We tend to spend the most money when shopping on weekends, particularly on Sundays between 4:00 and 6:00 P.M. Wednesday-morning shoppers spend the least. \(^{\star}\) Independent random samples of weekend and weekday shoppers were selected and the amount spent per trip to the mall was recorded as shown in the following table: $$\begin{array}{ll} \hline \text { Weekends } & \text { Weekdays } \\ \hline n_{1}=20 & n_{2}=20 \\ \bar{y}_{1}=\$ 78 & \bar{y}_{2}=\$ 67 \\ s_{1}=\$ 22 & s_{2}=\$ 20 \\ \hline \end{array}$$ a. Is there sufficient evidence to claim that there is a difference in the average amount spent per trip on weekends and weekdays? Use \(\alpha=.05\) b. What is the attained significance level?

Suppose that we wish to test the null hypothesis \(H_{0}\) that the proportion \(p\) of ledger sheets with errors is equal to .05 versus the alternative \(H_{a},\) that the proportion is larger than \(.05,\) by using the following scheme. Two ledger sheets are selected at random. If both are error free, we reject \(H_{0} .\) If one or more contains an error, we look at a third sheet. If the third sheet is error free, we reject \(H_{0}\). In all other cases, we accept \(H_{0}\). a. In terms of this problem, what is a type I error? b. What is the value of \(\alpha\) associated with this test? c. In terms of this problem, what is a type II error? d. Calculate \(\beta=P\) (type II error) as a function of \(p\)

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a Bernoulli- distributed population with parameter p. That is, $$ p\left(y_{i} | p\right)=p^{y_{1}}(1-p)^{1-y_{i}}, \quad y_{i}=0,1 $$ a. Suppose that we are interested in testing \(H_{0}: p=p_{0}\) versus \(H_{a}: p=p_{a},\) where \(p_{0}k^{*}\) for some constant \(k\) iii. Give the rejection region for the most powerful test of \(H_{0}\) versus \(H_{a}\) b. Recall that \(\sum_{i=1}^{n} Y_{i}\) has a binomial distribution with parameters \(n\) and \(p\). Indicate how to determine the values of any constants contained in the rejection region derived in part [a(iii)]. c. Is the test derived in part (a) uniformly most powerful for testing \(H_{0}: p=p_{0}\) versus \(H_{a}: p>p_{0} ?\) Why or why not?

What assumptions are made when a Student's \(t\) test is employed to test a hypothesis involving a population mean?

A manufacturer of hard safety hats for construction workers is concerned about the mean and the variation of the forces its helmets transmit to wearers when subjected to a standard external force. The manufacturer desires the mean force transmitted by helmets to be 800 pounds (or less), well under the legal 1000 -pound limit, and desires \(\sigma\) to be less than \(40 .\) Tests were run on a random sample of \(n=40\) helmets, and the sample mean and variance were found to be equal to 825 pounds and 2350 pounds \(^{2}\), respectively. a. If \(\mu=800\) and \(\sigma=40,\) is it likely that any helmet subjected to the standard external force will transmit a force to a wearer in excess of 1000 pounds? Explain. b. Do the data provide sufficient evidence to indicate that when subjected to the standard external force, the helmets transmit a mean force exceeding 800 pounds? c. Do the data provide sufficient evidence to indicate that \(\sigma\) exceeds \(40 ?\)

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