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Chapter 12: Problem 20

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(99 \%\) confidence, \(n=24\) c. \(90 \%\) confidence, \(n=13\)

Short Answer

Expert verified
The t critical values for each scenario are: a. 95% confidence, n=17: \(t_{critical} \approx 2.12\) b. 99% confidence, n=24: \(t_{critical} \approx 2.807\) c. 90% confidence, n=13: \(t_{critical} \approx 1.782\)

Step by step solution

01

Find Degrees of Freedom

In each case, we need to first find the degrees of freedom (df) by subtracting 1 from the sample size (n). a. n = 17, df = 17-1 = 16 b. n = 24, df = 24-1 = 23 c. n = 13, df = 13-1 = 12
02

Calculate the T Critical Values

For each case, we need to find the t critical value given the desired confidence level and the degrees of freedom. a. For a 95% confidence level and 16 degrees of freedom: The t critical value in the t-distribution table is calculated by finding the intersection of the row with 16 degrees of freedom and the column labeled with the 95% confidence level. Alternatively, we can use statistical software to get this value. \(t_{critical} \approx 2.12\) b. For a 99% confidence level and 23 degrees of freedom: The t critical value in the t-distribution table is calculated by finding the intersection of the row with 23 degrees of freedom and the column labeled with the 99% confidence level. Alternatively, we can use statistical software to get this value. \(t_{critical} \approx 2.807\) c. For a 90% confidence level and 12 degrees of freedom: The t critical value in the t-distribution table is calculated by finding the intersection of the row with 12 degrees of freedom and the column labeled with the 90% confidence level. Alternatively, we can use statistical software to get this value. \(t_{critical} \approx 1.782\) In summary, the t critical values for each scenario are: a. 95% confidence, n=17: \(t_{critical} \approx 2.12\) b. 99% confidence, n=24: \(t_{critical} \approx 2.807\) c. 90% confidence, n=13: \(t_{critical} \approx 1.782\)

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Key Concepts

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

T Critical Value
Understanding the t critical value is integral when calculating confidence intervals for a sample mean. It refers to the cutoff point on a t-distribution. To explain the t critical value in simpler terms, imagine a curve that outlines the probability of data points in a sample. For a given confidence level鈥攍ike 95% or 99%鈥攖he t critical value marks the boundary beyond which a certain percentage of the data points are expected to lie.

For instance, if you are working with a 95% confidence level, the t critical value tells us that we can expect 95% of sample means to fall within this interval if we were to draw an infinite number of samples from the population. The remaining 5% is split equally on both tails of the distribution, indicating the regions considered highly unlikely for the sample means to occur. To find this value, you can refer to a t-distribution table or use statistical software, based on the degrees of freedom for your data.
Degrees of Freedom
The concept of degrees of freedom (df) is like a tally of the number of values in a calculation that are free to vary. When calculating confidence intervals, degrees of freedom are critical because they determine the exact shape of the t-distribution you鈥檙e working with.

To put it simply, for a given sample size, subtract one to get your degrees of freedom (df = n - 1). Why subtract one? Because when estimating a population parameter like the mean, one value is lost to the sample mean calculation itself; it's this sample mean that constrains the system, leaving us with one less 'free' data point. This number then guides us in selecting the correct row in the t-distribution table for finding the t critical value.
T-Distribution
The t-distribution is a key player when we work with small sample sizes, especially when the population standard deviation is unknown. The shape of a t-distribution is similar to the normal distribution鈥攂ell-shaped and symmetric鈥攂ut with thicker tails. This shape means that there's a higher probability for values to fall further from the mean as compared to a normal distribution.

As sample size increases, the t-distribution gets closer to the normal distribution. This happens because, with more data, the estimate of the standard deviation becomes more reliable. When using the t-distribution to calculate the t critical value, remember that the specific form of the distribution is selected based on degrees of freedom, which correspond to the sample size.
Sample Size
Sample size is the number of observations or data points in a sample. It's denoted as 'n' and is an essential element for determining degrees of freedom and the shape of the t-distribution.

Why is sample size so important? Larger samples tend to more closely resemble the population from which they are drawn, providing a more accurate representation and allowing us to use the normal distribution as our model. However, with smaller samples, we must adjust our methods and rely on the t-distribution, as it accounts for the increased variability and uncertainty. The size of your sample also affects the width of the confidence interval: smaller samples generally lead to wider intervals, reflecting higher uncertainty around the estimate of the mean. It's a delicate balance; a too-small sample could lead to unreliable results, whereas an unnecessarily large sample might waste resources.

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Most popular questions from this chapter

Suppose that a random sample of 50 cans of a particular brand of fruit juice is selected, and the amount of juice (in ounces) in each of the cans is determined. Let \(\mu\) denote the mean amount of juice for the population of all cans of this brand. Suppose that this sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of (7.8,9.4) a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 7.8 and 9.4 . Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a random sample of size 50 and then calculating the corresponding \(95 \%\) confidence interval is repeated 100 times, exactly 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. a. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? Describe the shape of the sampling distribution of \(\bar{x}\). b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

Acrylic bone cement is sometimes used in hip and knee replacements to secure an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force has a distribution that is approximately normal, use a confidence interval to estimate the mean breaking force for acrylic bone cement.

The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald's restaurants in Europe in July 2016 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llll}4.44 & 3.15 & 2.42 & 3.96\end{array}\) \(\begin{array}{llll}4.51 & 4.17 & 3.69 & 4.62\end{array}\) \(\begin{array}{lll}3.80 & 3.36 & 3.85\end{array}\) The mean price of a Big Mac in the U.S. in July 2016 was \$5.04. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price? Test the relevant hypotheses using \(\alpha=0.05 .\) (Hint: See Example 12.12.)

Suppose that the population mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? At least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 \(\mathrm{mm}\) ? At least \(68 \mathrm{~mm} ?\)
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