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USA TODAY reported that the average amount of money spent on coffee drinks each month is \(\$ 78.00\) (USA Snapshot, November 4, 2016). a. Suppose that this estimate was based on a representative sample of 20 adult Americans. Would you recommend using the one-sample \(t\) confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans? Explain why or why not. b. If the sample size had been 200 , would you recommend using the one-sample \(t\) confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans? Explain why or why not.

Short Answer

Expert verified
a. We cannot conclusively recommend using the one-sample t confidence interval for the sample size of 20 adult Americans, as there is not enough information to determine if the underlying population is approximately normally distributed. b. For the sample size of 200 adult Americans, we would not recommend using the one-sample t confidence interval. Instead, we suggest using the one-sample z confidence interval because when the sample size is large enough, the Central Limit Theorem applies and the sampling distribution of the sample mean becomes approximately normal, regardless of the underlying population distribution.

Step by step solution

01

Determine given sample size

The problem states that the sample size in this case is 20 adult Americans. Therefore, we have n = 20.
02

Assess the suitability of the one-sample t confidence interval

Since the sample size is less than 30 and the population standard deviation is unknown, we can consider using the one-sample t confidence interval. However, before deciding, we should also verify if the underlying population is approximately normally distributed. The question does not provide enough information to determine the distribution of the population. Therefore, based on the provided information, we cannot conclusively recommend using the one-sample t confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans. ##Part b##
03

Determine given sample size

In this case, the problem states that the sample size is 200 adult Americans. Therefore, n = 200.
04

Assess the suitability of the one-sample t confidence interval

Since the sample size is now larger than 30 (n > 30), the one-sample t confidence interval is typically not recommended. For larger sample sizes, the one-sample z confidence interval is usually more appropriate. This is because when the sample size is large enough, the Central Limit Theorem applies, and the sampling distribution of the sample mean becomes approximately normal, regardless of the underlying population distribution. So, for this scenario with a sample size of 200, we would not recommend using the one-sample t confidence interval but rather suggest using the one-sample z confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans.

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Key Concepts

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

Confidence Interval
A confidence interval provides an estimated range of values which is likely to include an unknown population parameter. This statistical tool is vital when trying to infer population characteristics from sample data. Confidence intervals are constructed at a chosen confidence level, such as 95% or 99%, which tells us how confident we can be that this range contains the true population mean.

The width of a confidence interval gives us a visual sense of its accuracy; a narrower interval usually indicates more precise estimates. The one-sample t confidence interval is typically used when we deal with small samples (less than 30) and an unknown population standard deviation.
  • For small sample sizes, the distribution of the sample mean may not be normal, especially if the population distribution is unknown or skewed.
  • This requires us to use the t-distribution, which accounts for these additional uncertainties.
For larger samples, the sampling distribution becomes normal due to the Central Limit Theorem, allowing us to use the z-distribution instead.
Central Limit Theorem
The Central Limit Theorem is a fundamental concept in statistics that states, regardless of the distribution shape of the population, the distribution of the sample mean will approach a normal distribution as the sample size increases. This holds true even if the original variables themselves are not normally distributed.

This property is incredibly powerful as it allows statisticians to make inferences about the population even when they do not know the exact nature or shape of its distribution. The Central Limit Theorem comes into play especially when dealing with larger sample sizes (usually greater than 30). Hence, it justifies using the z-distribution for confidence intervals instead of the t-distribution in such cases.
  • The theorem provides the foundation for the practice of the one-sample z confidence interval for large samples.
  • The accuracy of estimations increases as the sample size grows, resulting in more reliable conclusions.
Thus, understanding the Central Limit Theorem is essential for effectively applying concepts like confidence intervals to real-world data.
Population Mean Estimation
Estimating the population mean is a common objective in statistics. It involves using a sample to infer the central tendency of the entire population. Given resource constraints, surveying an entire population is rarely feasible; thus, we rely on sample data.

Using a representative sample, we calculate a sample mean and use statistical methods to place this mean within a confidence interval. This process helps create an estimate that captures the true population mean with a certain level of confidence.
  • If we have a small sample size and unknown population standard deviation, a t-distribution is applied.
  • For larger samples or when the population variance is known, the z-distribution is more appropriate.
This way of estimating the population mean gives us statistically sound predictions about the broader population characteristics based solely on sample observations.

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Most popular questions from this chapter

The report "2016 Salary Survey Executive Summary" (National Association of Colleges and Employers, www.naceweb.org/uploadedfiles/files/2016/publications/executive- summary/2016-nace-salary-survey-fall-executive-summary. pdf, retrieved December 24,2016 ) states that the mean yearly salary offer for students graduating with mathematics and statistics degrees in 2016 was \(\$ 62,985 .\) Suppose that a random sample of 50 math and statistics graduates at a large university who received job offers resulted in a mean offer of \(\$ 63,500\) and a standard deviation of \(\$ 3300\). Do the sample data provide strong support for the claim that the mean salary offer for math and statistics graduates of this university is higher than the 2016 national average of \(\$ 62,985 ?\) Test the relevant hypotheses using \(\alpha=0.05\).

The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, \(\mu=0.5\) and \(\sigma=0.289\) a. If \(\bar{x}\) is the average waiting time for a random sample of \(n=16\) waiting times, what are the values of the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) b. Answer Part (a) for a random sample of 50 waiting times. Draw a picture of the approximate sampling distribution of \(\bar{x}\) when \(n=50\).

The report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

The paper "Alcohol Consumption, Sleep, and Academic Performance Among College Students" (Journal of Studies on Alcohol and Drugs [2009]: 355-363) describes a study of \(n=236\) students that were randomly selected from a list of students enrolled at a liberal arts college in the northeastern region of the United States. Each student in the sample responded to a number of questions about their sleep patterns. For these 236 students, the sample mean additional time spent sleeping on weekend days compared to the other days of the week was reported to be 1.29 hours and the standard deviation was 1.09 hours. Suppose that you are interested in learning about the value of \(\mu,\) the mean additional time spent sleeping on weekend days for students at this college. The following table is similar to the table that appears in Example 12.4 . The "what you know" information has been provided. Complete the table by filling in the "how you know it" column.

The article “The Association Between Television Viewing and Irregular Sleep Schedules Among Children Less Than 3 Years of Age" (Pediatrics [2005]: 851-856) reported the accompanying \(95 \%\) confidence intervals for average TV viewing time (in hours per day) for three different age groups. $$ \begin{array}{|lc|} \hline \text { Age Group } & \text { 95\% Confidence Interval } \\ \hline \text { Less than } 12 \text { Months } & (0.8,1.0) \\ 12 \text { to } 23 \text { Months } & (1.4,1.8) \\ 24 \text { to } 35 \text { Months } & (2.1,2.5) \\ \hline \end{array} $$ a. Suppose that the sample sizes for each of the three agegroup samples were equal. Based on the given confidence intervals, which of the age-group samples had the greatest variability in TV viewing time? Explain your choice. b. Now suppose that the sample standard deviations for the three age-group samples were equal, but that the three sample sizes might have been different. Which of the three age-group samples had the largest sample size? Explain your choice. c. The interval (0.7,1.1) is either a \(90 \%\) confidence interval or a \(99 \%\) confidence interval for the mean TV viewing time calculated using the sample data for children less than 12 months old. Is the confidence level for this inter- \(-\) val \(90 \%\) or \(99 \% ?\) Explain your choice.

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