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USA TODAY (October 14,2016 ) reported that Americans spend 4.1 hours per weekday checking work e-mail. This was an estimate based on a survey of 1004 white-collar workers in the United States. a. Suppose that you would like to know if there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8 -hour work day. What would you need to assume about the sample in order to use the given sample data to answer this question? b. Given that any concerns about the sample were satisfactorily addressed, carry out a test to decide if there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8 -hour work day. Suppose that the sample standard deviation was \(s=1.3\) hours.

Short Answer

Expert verified
In order to use the given sample data, we need to assume that the sample is random and representative of the entire population of US white-collar workers and that the time spent checking work e-mail is normally distributed, or the sample size is large enough to apply the Central Limit Theorem. Using a t-test, we find a test statistic of 2.44, which is greater than the critical value of 1.646 at a significance level of 0.05. Thus, we reject the null hypothesis and conclude that there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8-hour workday.

Step by step solution

01

Define the hypothesis test

We begin by defining our null and alternative hypotheses. For this problem, we are testing if the mean time spent checking work e-mail is more than half of the 8-hour workday, or 4 hours. We have: \(H_0: \mu = 4\) \(H_1: \mu > 4\) where \(\mu\) is the population mean time spent checking work e-mail.
02

Identify the assumptions

To use the sample data to perform the hypothesis test, we need to assume the following: 1. The sample is random and representative of the entire population of US white-collar workers. 2. The time spent checking work e-mail is normally distributed, or the sample size is large enough so that the Central Limit Theorem (CLT) holds true. The sample size is 1004, which is sufficiently large to apply CLT.
03

Identify the test statistic and compute its value

Since we don't know the population standard deviation, we will use the t-test statistic, which is computed as: \(t =\frac{(\bar{x} - \mu_0)}{(\frac{s}{\sqrt{n}})}\) where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. Using the given data: \(t =\frac{(4.1 - 4)}{(\frac{1.3}{\sqrt{1004}})}= \frac{0.1}{(0.041)} = 2.44\)
04

Determine the critical value and make a decision

Let's assume a significance level of \(\alpha=0.05\). Since we are conducting a right-tailed test, we want to determine the critical value of t (referred to as \(t_{\alpha}\)) at which we reject the null hypothesis when \(t > t_{\alpha}\). We find the critical t-value for \(\alpha=0.05\) and \(df=1003\), which is approximately 1.646. Since our calculated test statistic (2.44) is greater than the critical value (1.646), we reject the null hypothesis.
05

Interpretation of the result

We reject the null hypothesis and conclude that there is evidence that the mean time spent checking work e-mail for white-collar workers in the United States is more than half of the 8-hour workday at a 0.05 significance level.

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Key Concepts

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

Null and Alternative Hypothesis
Understanding the null and alternative hypothesis is a fundamental component of hypothesis testing in statistics. These hypotheses are competing claims about a population parameter, such as the mean or proportion. In the context of our exercise, the null hypothesis \(H_0\) asserts that the mean time white-collar workers in the United States spend checking work emails is equal to 4 hours, which represents half of an 8-hour workday. Symbolically, we represent this as \(H_0: \mu = 4\).

The alternative hypothesis \(H_1\), on the other hand, is what we're trying to find evidence for. In this case, it is the claim that the mean time is greater than 4 hours, denoted as \(H_1: \mu > 4\). To prove this, we collect sample data and conduct a test of significance. If the test provides sufficient evidence to support the alternative hypothesis, we can reject the null hypothesis and accept the alternative hypothesis as more likely. If not, we fail to reject the null hypothesis, meaning we don't have enough evidence to support the claim made by the alternative hypothesis.

The null hypothesis is often denoted as the 'status quo' and requires strong evidence to be rejected. On the other hand, the alternative hypothesis represents a new claim, suggesting a shift from the prevailing belief or established norm. The choice between a null and alternative hypothesis fundamentally shapes the direction and implications of the hypothesis test.
T-test Statistic
The t-test statistic is a crucial tool used in hypothesis testing when the population standard deviation is unknown and has to be estimated from the sample data. It allows us to determine how far the sample mean is from the hypothesized population mean in terms of standard errors. The formula to compute the t-test statistic is as follows:

\[t = \frac{(\bar{x} - \mu_0)}{(\frac{s}{\sqrt{n}})}\]

In this formula, \(\bar{x}\) represents the sample mean, \(\mu_0\) is the mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. The result is a t-value that lets us make an inference about the population mean.

In our exercise, the t-value obtained is 2.44. This number represents the number of standard errors the sample mean is above the hypothesized mean. We compare this t-value to a critical value from the t-distribution table corresponding to our chosen significance level and degrees of freedom (number of observations minus one). If our test statistic exceeds the critical value, we have sufficient evidence to reject the null hypothesis and conclude that the alternative hypothesis is more likely to be true, as was the case with our email checking time example.
Central Limit Theorem
The Central Limit Theorem (CLT) is a statistical theory that plays a pivotal role in the field of statistics, especially when dealing with sample means. The CLT states that given a sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately equal to the mean of the population. Furthermore, these sample means will be normally distributed, forming what is known as a sampling distribution, regardless of the shape of the population distribution.

The CLT allows us to perform hypothesis tests such as the t-test when we may not know the population's distribution. As long as the sample size is large enough, usually considered to be n > 30, we can rely on the CLT to assume that the sampling distribution of the mean will be normally distributed. This is essential for making inferences about the population mean using the sample mean and for calculating the test statistics accurately.

In the exercise provided, the sample size is 1004, which is well above 30, allowing us to apply the CLT with confidence. This underpins the validity of the t-test we conducted, even if the original population distribution of time spent checking emails is not normal. The central limit theorem thus provides the foundation for much of inferential statistics and is integral to hypothesis testing.

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Most popular questions from this chapter

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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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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