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Chapter 7: Problem 12

Basic Computation: Confidence Interval A random sample of size 81 has sample mean 20 and sample standard deviation 3 (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution to compute a confidence interval for the population mean \(\mu ?\) Explain. (b) Find a \(95 \%\) confidence interval for \(\mu\) (c) Interpretation Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
(a) No need for t-distribution; use z-distribution. (b) CI: (19.348, 20.652). (c) We are 95% confident that the true mean falls between 19.348 and 20.652.

Step by step solution

01

Check Requirements

We need to determine if it is appropriate to use a Student's \(t\) distribution. This is typically used when the population standard deviation is unknown and the sample size is small (usually \(n < 30\)). In this case, the sample size is 81, which is greater than 30, so a \(z\)-distribution could also be used. However, if the underlying population distribution is not known to be normal, using the \(t\)-distribution is generally safer. Since the sample size is large, we can use the \(z\)-distribution.
02

Calculate Standard Error

The standard error (SE) of the sample mean is calculated using the formula: \[SE = \frac{s}{\sqrt{n}}\]where \(s = 3\) is the sample standard deviation and \(n = 81\) is the sample size. Therefore,\[SE = \frac{3}{\sqrt{81}} = \frac{3}{9} = 0.333\]
03

Find Critical Value

For a 95% confidence interval, we need to find the critical value corresponding to the confidence level. If we use a \(z\)-distribution, the critical \(z\)-value is 1.96 for a two-tailed test. This value is derived from standard \(z\)-tables.
04

Compute Confidence Interval

The confidence interval is given by:\[\bar{x} \pm z^* \times SE\]where \(\bar{x} = 20\), \(z^* = 1.96\), and \(SE = 0.333\). Substituting in the values, \[20 \pm 1.96 \times 0.333 = 20 \pm 0.652\]So, the confidence interval is \[(19.348, 20.652)\]
05

Interpretation

A 95% confidence interval means that we can say with 95% confidence that the true population mean \(\mu\) lies between 19.348 and 20.652. This interval suggests the range within which the population mean is likely to fall.

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Key Concepts

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

Student's t-distribution
The Student's t-distribution is particularly useful when we deal with small sample sizes (typically less than 30) or when the population standard deviation is unknown. This distribution is bell-shaped and similar to the normal distribution but has thicker tails to account for the additional uncertainty in smaller samples.
The thicker tails mean there's a higher probability of values further away from the mean, which helps mitigate the risk of misestimating the mean of a small sample.
  • Best used when sample size is small and population standard deviation is unknown.
  • Has thicker tails than the normal distribution.
  • This makes it more adaptable to samples where extreme values might occur.

When having a larger sample size, such as 81 in our exercise, the Student's t-distribution becomes very similar to the standard normal distribution, making it less critical to differentiate between them.
z-distribution
The z-distribution, also known as the standard normal distribution, is a special case of the normal distribution where the mean is 0 and the standard deviation is 1. It is typically used in statistics when you have a large sample size, generally over 30.
In our exercise, the sample size is 81, which is considered large enough to use the z-distribution. This is because the larger the sample, the more the distribution of the sample mean will approximate a normal distribution due to the Central Limit Theorem.
  • Used primarily for large samples.
  • Mean is 0 and standard deviation is 1.
  • Assumes the population has a normal distribution or the sample size is sufficiently large.

Since the sample size in our exercise exceeded 30, both the Student's t-distribution and the z-distribution were options, but the z-distribution was chosen for simplicity and because the exact population distribution was unknown.
Critical Value
The critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis.
In the context of confidence intervals, the critical value defines the margin of error, dictating how wide the interval must be to encapsulate the true population parameter with a given level of confidence.
  • For a 95% confidence interval using the z-distribution, the critical value is 1.96.
  • The critical value is derived from z-tables or t-tables based on the confidence level.
  • Determines the span of the interval above and below the sample mean.

This means for our exercise, the critical value of 1.96 is used to calculate the 95% confidence interval for the population mean, ensuring that the mean lies within a certain range 95% of the time.
Standard Error
Standard Error (SE) is a measure of the variability or dispersion of the sample mean estimate of a population parameter. It helps assess how accurately the sample mean approximates the true population mean.
The standard error decreases as the sample size increases, indicating more precision in the sample's representation of the population.
  • Calculated by dividing the sample standard deviation by the square root of the sample size: \[SE = \frac{s}{\sqrt{n}}\]
  • Smaller SE means the sample mean is a more accurate estimation of the population mean.
  • In our exercise, with a sample size of 81 and standard deviation 3, SE is 0.333, indicating a reasonable degree of precision.

A smaller standard error, as calculated in our exercise, means our sample mean (20) is a reliable estimate of the population mean.

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Most popular questions from this chapter

Assume that the population of \(x\) values has an approximately normal distribution. Wildlife: Mountain Lions How much do wild mountain lions weigh? The 77 th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): \(68 \quad 104\) \(128 \quad 122 \quad 60\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x}=91.0\) pounds and \(s \approx 30.7\) pounds. (b) Find a \(75 \%\) confidence interval for the population average weight \(\mu\) of all adult mountain lions in the specified region. (c) Interpretation What does the confidence interval mean in the context of this problem?

What about the sample size \(n\) for confidence intervals for the difference of proportions \(p_{1}-p_{2} ?\) Let us make the following assumptions: equal sample sizes \(n=n_{1}=n_{2}\) and all four quantities \(n_{1} \hat{p}_{1}, n_{1} \hat{q}_{1}, n_{2} \hat{p}_{2},\) and \(n_{2} \hat{q}_{2}\) are greater than \(5 .\) Those readers familiar with algebra can use the procedure outlined in Problem 28 to show that if we have preliminary estimates \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and a given maximal margin of error \(E\) for a specified confidence level \(c,\) then the sample size \(n\) should be at least $$n=\left(\frac{z_{c}}{E}\right)^{2}\left(\hat{p}_{1} \hat{q}_{1}+\hat{p}_{2} \hat{q}_{2}\right)$$ However, if we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\), then the theory similar to that used in this section tells us that the sample size \(n\) should be at least $$n=\frac{1}{2}\left(\frac{z_{c}}{F}\right)^{2}$$ (a) In Problem 17 (Myers-Briggs personality type indicators in common for married couples), suppose we want to be \(99 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.04 .\) Use the preliminary estimates \(\hat{p}_{1}=289 / 375\) for the proportion of couples sharing two personality traits and \(\hat{p}_{2}=23 / 571\) for the proportion having no traits in common. How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )? (b) Suppose that in Problem 17 we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and we want to be \(95 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.05 .\) How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )?

At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of \(n_{1}=316\) patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of \(n_{2}=419\) patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. Let \(p_{1}\) be the population proportion of all patients with minor burns receiving the plasma compress treatment who have no visible scars. Let \(p_{2}\) be the population proportion of all patients with minor burns not receiving the plasma compress treatment who have no visible scars. (a) Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (c) Explain the meaning of the confidence interval found in part (b) in the context of the problem. Does the interval contain numbers that are all positive? all negative? both positive and negative? At the \(95 \%\) level of confidence, does treatment with plasma compresses seem to make a difference in the proportion of patients with visible scars from minor burns?

Answer true or false. Explain your answer. A larger sample size produces a longer confidence interval for \(\mu\).

A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p > 5\) and \(n q > 5\) since we usually do not know \(p,\) we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p}\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5 .\) Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed 5 Hint: In the inequality \(n \hat{p}>5,\) replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5,\) replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).
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