The number of concurrent users for some internet service provider has always averaged 5000 with a standard deviation of 700. After an equipment upgrade, the average number of users at 100 randomly selected moments of time is 5200. Does it indicate, at a 5% level of significance, that the mean number of concurrent users has increased? Assume that the standard deviation of the number of concurrent users has not changed.
Question
The number of concurrent users for some internet service provider has always averaged 5000 with a standard deviation of 700. After an equipment upgrade, the average number of users at 100 randomly selected moments of time is 5200. Does it indicate, at a 5% level of significance, that the mean number of concurrent users has increased? Assume that the standard deviation of the number of concurrent users has not changed.
Solution
To answer this question, we need to conduct a hypothesis test for the mean. Here are the steps:
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State the null and alternative hypotheses. The null hypothesis (H0) is that the mean number of concurrent users has not increased, i.e., the mean is still 5000. The alternative hypothesis (H1) is that the mean number of concurrent users has increased, i.e., the mean is greater than 5000.
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Calculate the test statistic. We use the formula for the z-score, which is (X̄ - μ) / (σ/√n), where X̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the given values, we get (5200 - 5000) / (700/√100) = 2.857.
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Determine the critical value. For a 5% level of significance and a one-tailed test, the critical value from the z-table is 1.645.
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Make the decision. Since the test statistic (2.857) is greater than the critical value (1.645), we reject the null hypothesis.
So, at a 5% level of significance, the data provides enough evidence to conclude that the mean number of concurrent users has increased.
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Example: The number of concurrent users for an ISP has historically averaged 5000. After a marketing campaign, the management would like to know if it has resulted in an increase in the number of concurrent users. To test this, data were collected by observing the number of concurrent users at 100 randomly selected moments of time. Suppose that the average and the standard deviation of the sample data are 5200 and 800, respectively. What is the null and alternative hypothesis if the study is to find for any evidence that the mean number of concurrent users has increased?
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A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours.A similar study conducted a year earlier estimated that μ, the mean number of weekly hours that U.S. adults use computers at home, was 8 hours. We would like to test (at the usual significance level of 5%) whether the current study provides significant evidence that this mean has changed since the previous year.Using a 95% confidence interval of (7.7, 9.3), which of the following is an appropriate conclusion? The current study does provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls outside the confidence interval. The current study does not provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls outside the confidence interval. The current study does provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls inside the confidence interval. The current study does not provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls inside the confidence interval. You cannot draw a conclusion because the only way to reach a conclusion is to find the p-value of the test.
A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours.A similar study conducted a year earlier estimated that μ, the mean number of weekly hours that U.S. adults use computers at home, was 8 hours. We would like to test (at the usual significance level of 5%) whether the current study provides significant evidence that this mean has changed since the previous year.Using a 95% confidence interval of (7.7, 9.3), which of the following is an appropriate conclusion? The current study does provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls outside the confidence interval. The current study does not provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls outside the confidence interval. The current study does provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls inside the confidence interval. The current study does not provide significant evidence that the mean number of weekly hours has changed over the past year, since 8 falls inside the confidence interval. You cannot draw a conclusion because the only way to reach a conclusion is to find the p-value of the test
A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours.We are 95% confident that the mean number of weekly hours that U.S. adults use computers at home falls between which of the following intervals? 6.5 and 10.5 7.7 and 9.3 8.4 and 8.6 8.1 and 8.9 7.3 and 9.7
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