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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

People in a random sample of 236 students enrolled at a liberal arts college were asked questions about how much sleep they get each night ("Alcohol Consumption, Sleep, and Academic Performance Among College Students," Journal of Studies on Alcohol and Drugs [2009]: 355-363). The sample mean sleep duration (average hours of daily sleep) was 7.71 hours and the sample standard deviation was 1.03 hours. The recommended number of hours of sleep for college-age students is 8.4 hours per day. Is there convincing evidence that the population mean daily sleep duration for students at this college is less than the recommended number of 8.4 hours? Test the relevant hypotheses using \(\alpha\) \(=0.01\)

A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]:1369-1374). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) in New York City were approached as they entered the restaurant at lunchtime and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburgerchain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2000 calories, the online Healthy Dining Finder (www.healthydiningfinder .com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=0.01\) d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not. e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased. f. Do you think that asking a customer to provide his or her receipt before they ordered could have introduced a bias? Explain.

The authors of the paper "Serum Zinc Levels of Cord Blood: Relation to Birth Weight and Gestational Period" (Journal of Trace Elements in Medicine and Biology [2015]: \(180-183)\) carried out a study of zinc levels of low-birth- weight babies and normal-birth-weight babies. For a sample of 50 lowbirth- weight babies, the sample mean zinc level was 17.00 and the standard error \(\left(\frac{s}{\sqrt{n}}\right)\) was \(0.43 .\) For a sample of 73 normal- birth-weight babies, the sample mean zinc level was 18.16 and the standard error was 0.32 . Explain why the two standard errors are not the same.

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 \(68 \mathrm{~mm} ?\)

What percentage of the time will a variable that has a distribution with the specified degrees of freedom fall in the indicated region? (Hint: See discussion on page \(581 .\) ) a. 10 df, between -1.81 and 1.81 b. 24 df, between -2.06 and 2.06 c. 24 df, outside the interval from -2.80 to 2.80 d. \(10 \mathrm{df}\), to the left of -1.81
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