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Chapter 9: Problem 29

Men's Pulse Rates (Example 10) A random sample of 25 men's resting pulse rates shows a mean of 72 beats per minute and a standard deviation of 13 . a. Find a \(95 \%\) confidence interval for the population mean pulse rate for men, and report it in a sentence. You may use the table given for Exercise \(9.25\). b. Find a \(99 \%\) confidence interval. c. Which interval is wider and why?

Short Answer

Expert verified
The 95% confidence interval for the population mean pulse rate for men is \((66.928, 77.072)\). The 99% confidence interval for the same is \((65.32, 78.68)\). The 99% confidence interval is wider than the 95% one because to be more confident that a population parameter is in a certain range, we have to allow greater uncertainty and thus a wider interval.

Step by step solution

01

Understanding Confidence Intervals

Confidence intervals provide a range of values, derived from a sample, that is likely to contain the value of an unknown population parameter. They are usually stated along with a confidence level which tells the degree of certainty that the parameter lies within the interval. The formula for confidence intervals is \(\overline{x} \pm (Z \cdot \frac{s}{\sqrt{n}})\) where \(\overline{x}\) is the sample mean, \(Z\) is Z-score (a measure of how many standard deviations an observation is from the mean), \(s\) is the sample standard deviation and \(n\) is the sample size.
02

Calculating 95% Confidence Interval

Using the provided data: mean \(\overline{x}\) = 72, standard deviation \(s\) = 13 and sample size \(n\) = 25, and the critical value for a 95% confidence interval \(Z\) = 1.96 (acquired from a standard Z-score table). The interval is calculated as \(72 \pm (1.96 \cdot \frac{13}{\sqrt{25}})\). After calculating numbers inside the parenthesis, the interval results in \(72 \pm 5.072\). Hence, the 95% confidence interval is \((66.928, 77.072)\). This means that we are 95% confident that the population mean is between 66.928 and 77.072.
03

Calculating 99% Confidence Interval

For calculating a 99% confidence interval, the critical value \(Z = 2.576\) (from the standard Z-score table). An interval is calculated same way as in step 2 but with a new Z value: \(72 \pm (2.576 \cdot \frac{13}{\sqrt{25}})\), resulting in \(72 \pm 6.68\). So the 99% confidence interval is \((65.32, 78.68)\). We are then 99% sure that the population mean falls between 65.32 and 78.68.
04

Comparing the Intervals

In step 2 we obtained the 95% confidence interval to be \((66.928, 77.072)\) and the 99% confidence interval obtained in step 3 was \((65.32, 78.68)\). Clearly, the 99% confidence interval is wider. This shows that increasing our confidence that the population parameter lies within the interval comes with a cost of uncertainty due to the wider range.

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Key Concepts

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

Population Mean
The population mean is a measure of central tendency, representing the average value of a population. It is what we're often trying to estimate using sample data.

In our example of men's pulse rates, the population mean would be the true average pulse rate for all men. We don't have this data, so we use a sample mean, calculated from a subset of the population, to make an educated guess.

This sample mean, in our example, is 72 beats per minute. It's important to recognize that this is just an estimate of the more extensive population's mean. By using sample data, we're trying to approximate what we cannot measure directly.
Sample Standard Deviation
Sample standard deviation is a statistic that measures the amount of variation or dispersion in a sample data set.

In simple terms, it tells us how much the pulse rates deviate from the average pulse rate of our sample. A smaller value would mean the pulse rates of most sampled men are close to the mean, while a larger value indicates more variation.

In the case of the men's pulse rate example, we have a standard deviation of 13. This means that on average, each pulse rate in our sample of 25 men can be found 13 beats per minute away from the mean of 72. Understanding this helps in assessing the reliability of our sample's mean as a representation of the population mean.
Z-score
A Z-score, part of the confidence interval formula, is a statistic that tells us how many standard deviations an element is from the mean of the dataset. It's used to standardize data, enabling us to use standard normal distribution tables.

It is critical in calculating confidence intervals because it provides the multiplier necessary to capture a certain confidence level. Different confidence levels, like 95% and 99%, have different Z-scores:
  • For a 95% confidence level, the Z-score is 1.96.
  • For a 99% confidence level, the Z-score is 2.576.
These Z-scores help define how wide our confidence interval will be. The larger the Z-score, the wider the confidence interval, indicating a higher certainty about containing the actual population mean.
Confidence Level
The confidence level represents the degree of certainty that the population parameter lies within the confidence interval.

It鈥檚 expressed as a percentage, such as 95% or 99%, that describes the proportion of intervals that will include the population mean if you were to repeat the study multiple times. In our pulse rate example:
  • At a 95% confidence level, we are confident that the true population mean is within the interval of (66.928, 77.072).
  • At a 99% confidence level, this interval becomes (65.32, 78.68), which is wider to reflect a higher confidence.
The trade-off with increasing confidence is a broader interval, as seen in our example. This larger interval incorporates more uncertainty, providing greater assurance but less precision.

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Most popular questions from this chapter

Vegetarians' Weights The mean weight of all 20-yearold women is 128 pounds (http://www.kidsgrowth.com). A random sample of 40 vegetarian women who are 20 years old showed a sample mean of 122 pounds with a standard deviation of 15 pounds. The women's measurements were independent of each other. a. Determine whether the mean weight for 20 -year old vegetarian women is significantly less than 128 , using a significance level of \(0.05\). b. Now suppose the sample consists of 100 vegetarian women who are 20 years old, and repeat the test. c. Explain what causes the difference between the \(\mathrm{p}\) -values for parts a and \(\mathrm{b}\).

Used Car Ages (Example 5) The mean age of all 638 used cars for sale in the Ventura Country Star one Saturday in 2013 was \(7.9\) years, with a standard deviation of \(7.7\) years. The distribution of ages is right-skewed. For a study to determine the reliability of classified ads, a reporter randomly selects 40 of these used cars and plans to visit each owner to inspect the cars. He finds that the mean age of the 40 cars he samples is \(8.2\) years and the standard deviation of those 40 cars is \(6.0\) years. a. Which of these four numerical values are parameters and which are statistics? b. \(\mu=? \sigma=? s=? \bar{x}=?\) c. Are the conditions for using the CLT fulfilled? What would be the shape of the approximate sampling distribution of a large number of means, each from a sample of 40 cars?

Exam Scores The distribution of the scores on a certain exam is \(N(70,10)\), which means that the exam scores are Normally distributed with a mean of 70 and standard deviation of \(10 .\) a. Sketch the curve and label, on the \(x\) -axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be bigger than 80\. Shade the region under the Normal curve whose area corresponds to this probability.

Babies Weights (Example 2) Some sources report that the weights of full-term newborn babies have a mean of 7 pounds and a standard deviation of \(0.6\) pound and are Normally distributed. a. What is the probability that one newborn baby will have a weight within \(0.6\) pound of the mean-that is, between \(6.4\) and \(7.6\) pounds, or within one standard deviation of the mean? b. What is the probability the average of four babies' weights will be within \(0.6\) pound of the mean; will be between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and \(\mathrm{b}\).

Tomatoes Use the data from Exercise \(9.36\). a. Using the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 5 pounds using a significance level of \(0.05\) ? Why or why not? The confidence interval is reported here: \(\mathrm{I}\) am \(95 \%\) confident the population mean is between \(4.9\) and \(5.3\) pounds. b. Now test the hypothesis that the population mean is not 5 pounds using the four step procedure. Use a significance level of \(0.05\) and number your steps.
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