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

Short Answer

Expert verified
When the sample standard deviation was given as 5 hours in part (a), the test did not provide evidence to reject the null hypothesis that the average time spent on the Internet by Canadians equals 12.5 hours. However, in part (b), where the standard deviation was lower (2 hours), the null hypothesis was rejected, providing evidence that Canadians spent more than 12.5 hours on the Internet on average. The smaller standard deviation in part (b) led to more sensitivity in the test, resulting in the rejection of the null hypothesis.

Step by step solution

01

Define the hypotheses

Start by stating the null and alternative hypotheses for both parts (a) and (b). The null hypothesis (H0) would be that the population mean equals 12.5 hours i.e., H0: µ = 12.5. The alternative hypothesis (Ha) would be that the population mean is greater than 12.5 hours, i.e., Ha: µ > 12.5.
02

Perform the hypothesis test for part (a)

Since the population standard deviation is not known and the sample size is large (n > 30), a Z test can be used. The Z test statistic is calculated as follows: Z = (X - µ) / (σ/√n). Here, X is the sample mean (12.7 hours), µ is the value from the null hypothesis (12.5 hours), σ is the sample standard deviation (5 hours) and n is the sample size (1000). Using these values in the formula gives a Z score. Then, compare this score to the critical Z score from the Z distribution table for a significance level of .05 (1.96 for a one-tailed test). If the calculated Z score > 1.96, then reject H0.
03

Perform the hypothesis test for part (b)

Repeat the same process as in Step 2, but this time with the sample standard deviation given as 2 hours.
04

Compare results

After computing the Z scores for both tests, compare them. If the Z score in part (b) is more significant than the critical value but the Z score from test (a) is not, this means H0 was rejected in part (b) but not in (a).
05

Explain difference in results

The key reason why H0 was rejected in part (b) but not (a) lies in the standard deviation. A smaller standard deviation means that the sample means are closer to the population mean. Thus, even small deviations from H0 will result in larger Z scores, making it easier to reject H0. Therefore, a lower standard deviation in test (b) resulted in the rejection of H0.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a core concept of hypothesis testing. It is a statement or assumption that there is no effect or no difference, and it is denoted as \(H_0\). In the context of our exercise on internet usage by Canadians, the null hypothesis assumed that the average time spent online is \(12.5\) hours per week, symbolically represented as \(H_0: \mu = 12.5\).

In many statistical tests, the null hypothesis is set up to be potentially rejected. This ensures that the test focuses on finding evidence against it. To determine if the null hypothesis holds true, researchers calculate a test statistic and compare it against critical values derived from a probability distribution.

Consider this a starting point for any hypothesis test. If there's significant evidence, you might reject \(H_0\), suggesting that an alternative statement provides a better explanation.
Alternative Hypothesis
The alternative hypothesis complements the null hypothesis. Denoted as \(H_a\), it suggests that there is an effect or a difference. In our exercise, the alternative hypothesis posited that the mean time Canadians spent online each week was greater than \(12.5\) hours, or \(H_a: \mu > 12.5\).

Contrary to the null hypothesis, the alternative hypothesis is what a researcher wants to prove. It reflects a change or a deviation from what's stated in the null hypothesis. In hypothesis testing, a significant result suggests that the evidence supports the alternative hypothesis.

It's worth noting that the formulation of the alternative hypothesis directly influences the directionality of the test. In our example, because it involves a claim about being 'greater than,' a one-tailed test was employed.
Z Test
A Z test is a statistical test used to determine if there is a significant difference between sample and population means. It's especially useful when the sample size is large (usually \(n > 30\)). This test calculates a Z score, which indicates how far and in what direction the sample mean deviates from the population mean in standard deviation units.

For our exercise, we calculated the Z score using the formula: \[ Z = \frac{X - \mu}{\sigma/\sqrt{n}} \] where \(X\) is the sample mean, \(\mu\) is the hypothesized population mean, \(\sigma\) is the sample standard deviation, and \(n\) is the sample size.

The computed Z score is then compared to the critical Z value, which is based on the chosen significance level. It provides a basis for deciding whether to reject the null hypothesis. In the setup given, if the calculated Z score exceeds the critical value (\(1.96\) for a one-tailed test at \(0.05\) significance level), we reject \(H_0\). This analysis helps determine whether observed data confirms or contradicts the general assumption about population means stated in the null hypothesis.
Significance Level
The significance level, often denoted by \(\alpha\), is a probability threshold set by the researcher, indicating how unlikely a result must be for it to be considered statistically significant. For example, a significance level of \(0.05\) means there's a 5% chance of rejecting the null hypothesis when it is actually true (Type I error).

In hypothesis testing, when the p-value of the test statistic (derived from the Z test in this context) is lower than or equal to \(\alpha\), we reject the null hypothesis. This rejection implies that the observed effect was unlikely to have occurred by random chance alone.

The choice of significance level is critical. It reflects the risk the researcher is willing to take regarding false positives. In the provided exercise scenario, a \(0.05\) significance level was chosen, a common practice in many fields. This means researchers accepted a moderate level of skepticism about their findings, ensuring that if they detect a significant effect, it's likely robust.

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Most popular questions from this chapter

Suppose that you are an inspector for the Fish and Game Department and that you are given the task of determining whether to prohibit fishing along part of the Oregon coast. You will close an area to fishing if it is determined that fish in that region have an unacceptably high mercury content. a. Assuming that a mercury concentration of \(5 \mathrm{ppm}\) is considered the maximum safe concentration, which of the following pairs of hypotheses would you test: $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu>5 $$ or $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu<5 $$ Give the reasons for your choice. b. Would you prefer a significance level of \(.1\) or \(.01\) for your test? Explain.

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To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed 100 \(\mathrm{lb} / \mathrm{in} .^{2} .\) The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

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