Question

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 sample 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. c. Explain why the null hypothesis was rejected in the test of Part (b) but not in the test of Part (a).

Short answer

The null hypothesis asserting that the mean time Canadians spend online is equal to 12.5 hours cannot be rejected with a sample standard deviation of 5 hours but can be rejected when the sample standard deviation is reduced to 2 hours.

Step-by-step solution

Identify Null and Alternative Hypothesis

The null hypothesis is \(H_0: \mu = 12.5\) and the alternative hypothesis is \(H_a: \mu > 12.5\). Here \(\mu\) represents the mean time spent by Canadians using the Internet.

Part a: Perform Hypothesis Testing with Standard Deviation of 5

Since the standard deviation is given, we'll conduct a t-test. Firstly, calculate the t-value using formula:\n\(t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}\)\nwhere \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under null hypothesis, \(s\) is the sample standard deviation and \(n\) is the number of observations. Substituting the values, we get \(t = \frac{12.7 - 12.5}{5/\sqrt{1000}} = 1.778\).\nUsing the t-distribution table, we find the area to the right of 1.778 (our p-value). If this p-value is less than the significance level of 0.05, we reject the null hypothesis. In our case, the p-value is greater than 0.05, implying we don't have enough evidence to reject the null hypothesis.

Part b: Perform Hypothesis Testing with Standard Deviation of 2

Let's redo the calculations with a standard deviation of 2. The t-value now becomes \(\frac{12.7 - 12.5}{2/\sqrt{1000}} = 4.472\). Once again, find the area to the right of the t-value from the t-table. The p-value in this case will be much smaller than in the part a, and also less than 0.05, indicating strong evidence against the null hypothesis. Hence, the null hypothesis will be rejected in this case.

Explain the Difference

The difference in results in part a and part b arises from the different standard deviations. A smaller standard deviation means the sample mean is a more reliable estimate of the population mean. As standard deviation decreases, the t-value increases and the p-value decreases, making it more likely to reject the null hypothesis.

Key concepts

Understanding the t-test

The t-test is a statistical method used to determine whether there is a significant difference between the means of two groups. In our exercise, we used the t-test to test if the mean time Canadians spend on the Internet is greater than 12.5 hours. The formula for the t-test is: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where:

• \( \bar{x} \) is the sample mean,
• \( \mu_0 \) is the mean under the null hypothesis,
• \( s \) is the sample standard deviation,
• \( n \) is the number of observations.

This test helps in understanding if the observed data deviates significantly from the null hypothesis, which assumes no effect or no difference.

The Null Hypothesis

In hypothesis testing, the null hypothesis (\(H_0\)) is a statement that there is no effect or no difference. It is the hypothesis that we aim to test against using sample data. For our study, the null hypothesis is that the mean time spent online by Canadians is equal to 12.5 hours: \[ H_0: \mu = 12.5 \] Rejecting the null hypothesis suggests that there is enough evidence to support the alternative hypothesis, which in this case indicates that the mean time is greater than 12.5 hours. If the null hypothesis is not rejected, we conclude that the sample data did not provide sufficient evidence to say otherwise. Thus, the null hypothesis remains valid.

Role of Standard Deviation

Standard deviation is a crucial concept in statistics, representing how much individual data points deviate from the mean. It provides insight into the dispersion of the dataset. A lower standard deviation indicates that the data points are close to the mean, while a higher standard deviation implies that they are spread out over a wider range. In our exercise:

• A standard deviation of 5 resulted in a t-value that was not significant enough to reject the null hypothesis.
• A smaller standard deviation of 2 led to a larger t-value, which resulted in rejecting the null hypothesis.

This demonstrates that the smaller the standard deviation, the more likely we are to detect a significant difference if one exists. The precision of our mean estimate improves, thus increasing the chances of rejecting a false null hypothesis.

Significance Level in Hypothesis Testing

The significance level, often denoted by \( \alpha \), is the threshold at which we decide whether a null hypothesis should be rejected. It represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. Common significance levels used in hypothesis testing are 0.01, 0.05, and 0.10. In our case, the significance level was set at 0.05. This means that there is a 5% risk of concluding that there is a difference when there is none.

• If the \( p \)-value is less than \( \alpha \), the result is statistically significant, and we reject the null hypothesis.
• If the \( p \)-value is greater than \( \alpha \), we do not reject the null hypothesis, indicating that there is not enough evidence to support a significant difference.

Selecting an appropriate significance level is essential, as it balances the risk of making Type I errors with the goal of making valid inferences from the data.