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For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. 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. Pulse Rates of Identical Twins A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eight sets of twins were randomly selected. The rates are given in the table as number of beats per minute. At \(\alpha=0.01,\) is there a significant difference in the average pulse rates of twins? Use the \(P\) -value method. Find the \(99 \%\) confidence interval for the difference of the two. $$ \begin{array}{l|cccccccc}{\text { Twin } \mathbf{A}} & {87} & {92} & {78} & {83} & {88} & {90} & {84} & {93} \\ \hline \text { Twin B } & {83} & {95} & {79} & {83} & {86} & {93} & {80} & {86}\end{array} $$

Short Answer

Expert verified
No significant difference at \( \alpha=0.01 \); 99% CI: (-3.43, 5.93).

Step by step solution

01

State Hypotheses and Identify the Claim

We are interested in determining if there is a significant difference in the average pulse rates of identical twins. We can set up the null hypothesis as follows:- Null Hypothesis \( H_0: \mu_d = 0 \), meaning there is no difference in the mean pulse rates of twins A and B.- Alternative Hypothesis \( H_1: \mu_d eq 0 \), indicating a difference in the mean pulse rates. The claim is the alternative hypothesis that there is a significant difference.
02

Find the Critical Value

The level of significance is \( \alpha = 0.01 \). Since this is a two-tailed test, we need to find the critical values for a t-distribution with \( n - 1 = 7 \) degrees of freedom. Using a t-distribution table or calculator, the critical values are approximately \( t_{\alpha/2} = \pm 3.499 \).
03

Compute the Test Value

Next, we calculate the test statistic using the paired sample t-test formula:\[ t = \frac{\bar{d}}{s_d/\sqrt{n}}\]Where \( \bar{d} \) is the mean of the differences between paired samples, \( s_d \) is the standard deviation of the differences, and \( n \) is the number of pairs.Calculate the differences: \[87-83=4, \ 92-95=-3, \ 78-79=-1, \ 83-83=0, \ 88-86=2, \ 90-93=-3, \ 84-80=4, \ 93-86=7\]Calculate mean of differences: \( \bar{d} = \frac{10}{8} = 1.25 \)Find the standard deviation: \( s_d = 3.776 \)Now plug these into the formula:\[ t = \frac{1.25}{3.776/\sqrt{8}} \approx 1.177 \]
04

Make the Decision

Compare the test statistic with the critical value. The calculated test statistic \( t = 1.177 \) does not fall in the rejection region, which has the critical values \( \pm 3.499 \). Therefore, we fail to reject the null hypothesis.
05

Summarize the Results

Since we failed to reject the null hypothesis, we conclude that there is not enough evidence at the \( \alpha = 0.01 \) level to support the claim that there is a significant difference in the average pulse rates of twins.
06

Calculate the Confidence Interval

Calculate the 99% confidence interval for the difference in means using:\[ \bar{d} \pm t_{\alpha/2} \times \frac{s_d}{\sqrt{n}}\]Using the critical value \( t_{\alpha/2} = 3.499 \), mean difference \( \bar{d} = 1.25 \), and standard error \( \frac{3.776}{\sqrt{8}} = 1.334 \):\[ 1.25 \pm 3.499 \times 1.334 = ( -3.43, 5.93 )\]This means that with 99% confidence, the true mean difference between the pulse rates of twins A and B is between -3.43 and 5.93 beats per minute.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a foundational concept in statistics. It represents a statement of no effect or no difference, serving as a baseline to be tested against.
In the case of comparing the pulse rates of identical twins, the null hypothesis (denoted as \( H_0 \)) states there is no difference in the mean pulse rates between twin A and twin B: \( \mu_d = 0 \).
This implies that any difference observed between the twins' pulse rates is merely due to random chance rather than a real effect. It is essential to note that in hypothesis testing, we never prove the null hypothesis true; we only gather evidence to reject or fail to reject it.
When the null hypothesis is tested, it's typically against an alternative hypothesis, which in this case, suggests there is a significant difference in the mean pulse rates \( \mu_d eq 0 \). Identifying the null hypothesis accurately helps in determining the direction and nature of our statistical test.
T-Distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution, but with thicker tails. This distribution is particularly useful when dealing with small sample sizes where the population standard deviation is unknown.
In our exercise, the data on pulse rates consists of only 8 pairs, which is a small sample. Accordingly, the t-distribution is employed because it accommodates the additional uncertainty inherent in smaller samples better than the normal distribution.
The critical value for the test is obtained from the t-distribution, particularly the value \( t_{\alpha/2} \), which defines the threshold beyond which we would consider the result statistically significant.
The degrees of freedom, which for a paired sample t-test is \( n - 1 \), where \( n \) is the number of pairs (in this case, 7), play a vital role in finding this critical value.
Thus, by having the proper understanding of the t-distribution, one can correctly interpret and implement hypothesis tests for small sample sizes.
Confidence Interval
A confidence interval provides a range of values that is believed to contain the true population parameter (such as the mean difference in pulse rates, in our exercise) with a certain level of confidence.
A 99% confidence interval is particularly stringent, offering high assurance that the interval captured the true parameter.
This interval is calculated to determine if the difference between the corresponding means is statistically significant at a given confidence level.
In the exercise, the confidence interval for the difference in pulse rates between twins is calculated using the formula \( \bar{d} \pm t_{\alpha/2} \times \frac{s_d}{\sqrt{n}} \), where \( \bar{d} \) is the mean of the differences, and \( s_d \) is the standard deviation of these differences.
The resulting interval from our exercise was (-3.43, 5.93), suggesting that with 99% confidence, the real difference in pulse rates lies somewhere in this range. If `0` is included in this interval, it implies that the difference could be non-significant, which aligns with failing to reject the null hypothesis.
Paired Sample T-Test
The paired sample t-test is a statistical method used to compare the means from two related groups. This test is instrumental when you have two measurements from the same group or individual at different times or under different conditions.
In our exercise, each twin's pulse rate is compared against the other's in sets, making a paired sample t-test appropriate. It helps in examining if there's a significant mean difference in pulse rates from these paired observations.
The test statistic is calculated using \( t = \frac{\bar{d}}{s_d/\sqrt{n}} \), where \( \bar{d} \) is the mean difference, and \( s_d \) is the standard deviation of the differences.
By using the paired sample t-test, one accounts for variability both within and across pairs, enhancing the precision of the estimate of mean difference compared to other tests that treat groups as independent.
This test concluded that no significant difference exists based on the calculated statistics, a critical outcome for the hypothesis regarding twin pulse rates.

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Most popular questions from this chapter

For Exercises 7 through \(27,\) 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. Hypertension 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 women are hypertensive. 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 43 patients had high blood pressure Women 52 patients had high blood pressure

Find each \(X,\) given \(\hat{p}\) a. \(\hat{p}=0.60, n=240\) b. \(\hat{p}=0.20, n=320\) c. \(\hat{p}=0.60, n=520\) d. \(\hat{p}=0.80, n=50\) e. \(\hat{p}=0.35, n=200\)

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. 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 and assume the variances are unequal. Waterfall Heights Is there a significant difference at \(\alpha=0.10\) in the mean heights in feet of waterfalls in Europe and the ones in Asia? The data are shown. $$ \begin{array}{ccc|ccc}{} & {} & {\text { Europe }} & {} & {\text { Asia }} \\\ \hline 487 & {1246} & {1385} & {614} & {722} & {964} \\ {470} & {1312} & {984} & {1137} & {320} & {830} \\ {900} & {345} & {820} & {350} & {722} & {1904}\end{array} $$

Instead of finding the mean of the differences between \(X_{1}\) and \(X_{2}\) by subtracting \(X_{1}-X_{2},\) you can find it by finding the means of \(X_{1}\) and \(X_{2}\) and then subtracting the means. Show that these two procedures will yield the same results.

For Exercises 7 through \(27,\) 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. Airlines On-Time Arrivals The percentages of on-time 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? $$ \begin{array}{ccc}{} & {\text { Airline } \mathbf{A}} & {\text { Airline } \mathbf{B}} \\ \hline \text { No.of flights } & {300} & {250} \\ {\text { No. of on-time flights }} & {213} & {185}\end{array} $$

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