91Ó°ÊÓ

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the 1000 resulting observations was \(12.7\) hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of \(.05\) to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of . 05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the mean time Canadians spend on the internet, \(\mu\), is equal to 12.5 hours. The alternative hypothesis (\(H_1\)) is that the mean time \(\mu\) is greater than 12.5 hours.

02

Formulate an Analysis Plan

We will perform a one-sample t-test. For this test, the test statistic is given by \(T = (\bar{x} - \mu_0) / (s/\sqrt{n})\) where \(\mu_0 = 12.5\) hours, \(s\) is the sample standard deviation, \(\bar{x} = 12.7\) hours is the sample mean, and \(n = 1000\) is the number of observations. The critical value for the test will be determined using the t-distribution with \( n - 1 = 999\) degrees of freedom and a significance level of \(0.05\). We will reject \(H_0\) if the computed T is greater than the critical value.

03

Analyze Sample Data - Case of 5 hours

Let's calculate the T statistic for the case where the standard deviation is 5 hours. The formula is \(T = (12.7 - 12.5) / (5/\sqrt{1000}) = 0.2 / (5/31.62) \approx 1.26\). The critical value for a t-distribution with 999 degrees of freedom and a significance level of 0.05 is approximately 1.646. Because 1.26 is less than 1.646, we will not reject the null hypothesis for this case.

04

Interpret the Results - Case of 5 hours

Since the T statistic is less than the critical value, we do not reject the null hypothesis. There is no sufficient evidence to conclude that Canadians spend more than 12.5 hours, on average, on the internet per week when the standard deviation is 5 hours.

05

Analyze Sample Data - Case of 2 hours

Now, let's calculate the T statistic for the case where the standard deviation is 2 hours. The formula is \(T = (12.7 - 12.5) / (2/\sqrt{1000}) = 0.2 / (2/31.62) \approx 3.16 \). Because 3.16 is greater than the critical value of 1.646, we reject the null hypothesis for this case.

06

Interpret the Results - Case of 2 hours

Since the T statistic is greater than the critical value, we reject the null hypothesis. There is sufficient evidence to conclude that Canadians spend more than 12.5 hours, on average, on the internet per week when the standard deviation is 2 hours.

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Key Concepts

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

t-distribution

The t-distribution is a crucial concept in statistics, especially when dealing with small sample sizes or when the population standard deviation is not known. It resembles a standard normal distribution but has heavier tails, allowing for greater variability in data. This is why it comes in handy for hypothesis testing.
When you perform a hypothesis test for the mean of a population, and the sample size is not large, or the population standard deviation is unknown, you rely on the t-distribution.

• The shape of the t-distribution depends on the degrees of freedom, which in simple cases is calculated as the sample size minus one.
• As the sample size increases, the t-distribution approaches a normal distribution.
• This distribution provides a critical value against which the calculated t-statistic is compared to determine if you should reject or not reject the null hypothesis.

Understanding the t-distribution is key for evaluating the significance of your test results, especially in cases where variability might skew outcomes, as seen in the example with varying standard deviations.

null hypothesis

In hypothesis testing, the null hypothesis ( \( H_0 \) ) is the statement that there is no effect or no difference. It is a starting assumption that an investigator attempts to test. For example, in the given scenario, our null hypothesis was that the mean time Canadians spend on the internet, denoted as \( \mu \), was equal to 12.5 hours.
The purpose of the null hypothesis is to provide a baseline which you can use to evaluate your actual observations against.

• Establishing the null hypothesis helps to organize the analysis and formulate conclusions based on statistical evidence.
• Deciding whether to reject the null hypothesis is done by comparing a calculated test statistic with a critical value derived from the t-distribution.
• Failing to reject the null doesn't mean it's true, it means we don't have sufficient evidence against it.

Successfully grasping the concept of null hypothesis helps in structuring and framing the entire hypothesis testing process.

sample standard deviation

The sample standard deviation is a measure of the amount of variation or dispersion in a set of values. It is a critical component in the calculation of the t-statistic, which is used in the hypothesis testing process.
In essence, the sample standard deviation helps us understand how spread out the data is from the mean. In hypothesis testing, smaller standard deviations can provide stronger evidence against the null hypothesis because the data points are closer to the mean.

• The sample standard deviation is used to calculate the standard error of the mean, which is important for determining the t-statistic.
• As shown in the steps above, when the sample standard deviation changed from 5 hours to 2 hours, the outcome of the hypothesis test changed.
• It is calculated as the square root of the variance, which is the average squared deviation of each number from the mean.

Recognizing how the sample standard deviation influences hypothesis test results is pivotal, as it can impact whether or not the null hypothesis is rejected, affecting your conclusion about the population mean.