91Ó°ÊÓ

How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hour. Source: American Time Use Survey conducted by the Bureau of Labor Statistics (a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day. (b) There are over 200 million Americans age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(95 \%\) confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. (d) Could the interval be used to estimate the mean amount of time a 9-year- old American spends eating and drinking each day? Explain.

Step by step solution

01

Explain the need for large sample size

A large sample size is needed because the histogram of time spent eating and drinking is skewed right. In a skewed distribution, individual samples may not be representative of the population mean. A larger sample size ensures the Central Limit Theorem applies, making the sampling distribution of the sample mean approximately normal.

02

Satisfy the requirements for constructing a confidence interval

The requirements for constructing a confidence interval are: (1) the sample is random, and (2) the sample size is large enough (typically n > 30) or the population is normal. Since the sample of 1001 Americans is random and the sample size is larger than 30, these conditions are satisfied.

03

Calculate the confidence interval

First, identify the sample mean \(\bar{x} = 1.22\) hours, standard deviation \(s = 0.65\) hours, and sample size \(n = 1001\). The critical value for a 95% confidence interval is approximately 1.96. Compute the margin of error (ME) using the formula: \[ ME = 1.96 * \frac{s}{\sqrt{n}} \] \[ ME = 1.96 * \frac{0.65}{\sqrt{1001}} \] \[ ME \approx 0.04 \] Finally, calculate the confidence interval: \[ CI = \bar{x} \pm ME \] \[ CI = 1.22 \pm 0.04 \] Therefore, the confidence interval is \[ (1.18, 1.26) \]

04

Interpret the confidence interval

The 95% confidence interval (1.18, 1.26) means we are 95% confident that the true mean amount of time Americans age 15 or older spend eating and drinking each day lies between 1.18 and 1.26 hours.

05

Determine if the interval applies to 9-year-olds

The interval cannot be used to estimate the mean time a 9-year-old spends eating and drinking because the data is specific to Americans age 15 or older. The behavior and habits of younger children, like a 9-year-old, may differ significantly.

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Key Concepts

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

Sample Size

When dealing with statistical studies, the size of the sample is crucial. In this exercise, a large sample size of 1001 Americans is chosen. A larger sample gives more reliable results because it represents the population better.
This is especially important in skewed distributions, where one side of the data (skewed right, in this case) might dominate.
Larger samples help balance out these extremes, making our results more accurate. This is why the Central Limit Theorem, which we will discuss soon, can apply effectively.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental concept in statistics. It states that the distribution of the sample mean will approach a normal distribution, regardless of the original population's distribution, given a large enough sample size.
In simpler terms, if we average enough random samples, the result will look like a bell-shaped curve, or normal distribution. This makes it easier to apply statistical methods like confidence intervals.
In this exercise, despite the skewed right distribution of the data, the large sample size allows us to apply the CLT and accurately construct the confidence interval.

Margin of Error

The margin of error (ME) quantifies the uncertainty in our sample mean estimate. It provides a range around our sample mean within which the true population mean is likely to fall.
To calculate the ME, we use the formula:
\[ ME = z * \frac{s}{\sqrt{n}} \]
Here, \(z\) is the critical value corresponding to our desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.
In this exercise, we use a 95% confidence level, so the critical value is 1.96. Calculations show the ME as approximately 0.04, giving the confidence interval.

Random Sampling

Random sampling is vital to ensure that each individual has an equal chance of being chosen. This minimizes bias and makes the sample representative of the population.
In our exercise, the random sample of 1001 Americans helps satisfy one of the key requirements for constructing a confidence interval, ensuring our results are generalizable to all Americans aged 15 or older.
Random sampling lays the foundation for reliable and valid statistical findings.

Skewed Distribution

A skewed distribution means that the data is not symmetrical. In this exercise, the time Americans spend eating and drinking each day is skewed right.
This means there are more instances of lower values, but some very high values shift the mean to the right.
Such skewness makes it challenging to estimate the mean accurately with small samples.
To overcome this, we use a larger sample size and rely on the Central Limit Theorem, which allows us to apply normal distribution methods, like constructing confidence intervals.