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?

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.

A lunch bar has an average sales amount of $8.72.  The manager initiates staff training to try to get them to sell more extras (chips, drinks etc), and after this training, a sample of 57 customers had the following sample statistics:9.28 and s = 2.36Assume our question of interest is: "is there significant evidence that the mean amount spent by customers has increased?"  Calculate the observed  value of the test statistic (correct to 2dp).

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.

Netflix: A study conducted by a technology company showed that the mean time spent per day browsing the video streaming service Netflix for something to watch was 19.9 minutes. Assume the standard deviation is =σ6. Suppose a simple random sample of 103 visits taken this year has a sample mean of =x20.9 minutes. A social scientist is interested to know whether the mean time browsing Netflix has increased. Use the =α0.01 level of significance and the P-value method with the TI-84 Plus calculator.Part 1 of 5(a) State the appropriate null and alternate hypotheses.H0: =μ19.9H1: >μ19.9This hypothesis test is a ▼right-tailed test.Part: 1 / 51 of 5 Parts CompletePart 2 of 5(b) Compute the value of the test statistic. Round the answer to two decimal places.z=

A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a mean different from the 600 mg claimed by the manufacturer. Test this claim at the p < .05 level of significance. The mean acetaminophen content for a random sample of n = 41 tablets is 603.3 mg. Assume that the population standard deviation is 4.9 mg.Question 12Answera.Since the test statistic is greater than the critical z, there is sufficient evidence to reject thenull hypothesis and to support the claim that the mean content of acetaminophen is not600 mg.b.Since the test statistic is greater than the critical z, there is insufficient evidence to rejectthe null hypothesis and to support the claim that the mean content of acetaminophen isnot 600 mgc.Since the test statistic is greater than the critical z, there is sufficient evidence to acceptthe null hypothesis and to support the claim that the mean content of acetaminophen is600 mg.d.Since the test statistic is less than the critical z, there is sufficient evidence to reject thenull hypothesis and to support the claim that the mean content of acetaminophen is not600 mg.