According to Facebook’s self-reported statistics, the average Facebook user is connected to 80 community pages, groups, and events. For a statistics project, a student at Contra Costa College tests the hypothesis that CCC students will average less than 80 such connections.She posts a survey on her Facebook page. Her sample contains 45 responses.She chooses a 5% level of significance. From her data she calculates a t-test statistic of approximately −1.74 with a P-value of about 0.04.What can she conclude? Nothing. The conditions for use of a t-test are not met. She cannot trust that the P-value is accurate for this reason. The data is statistically significant. In other words, the data do provide enough evidence to conclude that the mean number of Facebook connections to community pages, groups and events of all all CCC college students is less the 80. The data is not statistically significant. In other words, the data do not provide enough evidence to conclude that the mean number of Facebook connections to community pages, groups and events of all CCC college students is less the 80.

According to Facebook’s self-reported statistics, the average Facebook user has 130 Facebook friends. For a statistics project, a student at Contra Costa College tests the hypothesis that CCC students will average more than 130 Facebook friends.She randomly selects 3 classes from the schedule of classes and distributes a survey in these classes. Her sample contains 45 students.Here are the null and alternative hypotheses for her study: H0: µ = 130, Ha: µ > 130.From her survey data, the statistics student calculates that the mean number of Facebook friends for her sample is 138.7 with a standard deviation of 79.3. She analyzed her data using a t-test and obtained a P-value of 0.23.What conclusion can she draw from her data? Even though 138.7 is larger than 130, it is not significantly larger than 130. In other words, the data do not provide enough evidence to conclude that the mean number of Facebook friends of all CCC college students is higher than 130. The sample value of 138.7 is significantly larger than 130. In other words the data provide provide enough evidence to conclude that the mean number of Facebook friends of all CCC college students is higher than 130. Nothing. The conditions for use of a t-test were not met. She cannot trust that the P-value is accurate for this reason.

A student in a statistics class needs to do a project and decides to find out if the average high school GPA of students in a statistics course at her university is higher than the average high school GPA for their university. The university’s average high school GPA of enrolled students in their year was 4.41.She emails a survey to all of her friends that are taking statistics and asks them what their high school GPA is. Her sample contains 18 responses. She chooses a 5% level of significance. From her data, she calculates a t-test statistic of approximately 1.41 with a p-value of about 0.08. What can she conclude? The data is not statistically significant. In other words, the data do not provide enough evidence to conclude that the mean GPA of students taking statistics at her university is higher than 4.41. The data is statistically significant. In other words, the data do provide enough evidence to conclude that the mean GPA of students taking statistics at her university is higher than 4.41. Nothing. The conditions for use of a t-test are not met. She cannot trust that the p-value is accurate for this reason.

According to a Pew Research Center, in May 2011, 35% of all American adults had a smartphone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students.She selects 300 community college students at random and finds that 120 of them have a smartphone. In testing the hypotheses H0, P = 0.35, versus Ha, p > 0.35, she calculates the test statistic as Z = 1.82.Use the normal table to identify the appropriate p-value for this Z score.Click here to access the normal table.Given these results, which of the following is an appropriate conclusion? There is enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.034). There is enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.068). There is not enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.966). There is not enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.034).

In 2013, the average Girl Scout in New York City sold 96 boxes of cookies. The leader of Troop 5078 in New York City wants to know if the scouts in her troop sold more cookies than the average in New York City. She randomly samples 50 girls in Troop 5078 and records the number of boxes of cookies sold for each girl in the sample.The troop leader finds that her Girl Scouts each sold an average of 101.1 boxes of cookies with a standard deviation of 29.3. She analyzed her data using a t-test and obtained a p-value of 0.11.What conclusion can she draw from her data? The sample value of 101.1 is significantly larger than 96. In other words, the data provide enough evidence to conclude that the mean number of boxes of cookies sold by Girl Scouts in Troop 5078 was higher than 96. Even though 101.1 is larger than 96, it is not significantly larger than 96. In other words, the data do not provide enough evidence to conclude that the mean number of boxes of cookies sold by Girl Scouts in Troop 5078 was higher than 96. Nothing. The conditions for use of a t-test were not met. She cannot trust that the p-value is accurate for this reason.

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 20.6 minutes. Assume the standard deviation is =σ8. Suppose a simple random sample of 102 visits taken this year has a sample mean of =x19.6 minutes. A social scientist is interested to know whether the mean time browsing Netflix has decreased. Use the =α0.05 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: =μ20.6H1: <μ20.6This hypothesis test is a ▼left-tailed test.Part 2 of 5(b) Compute the value of the test statistic. Round the answer to two decimal places.z=−1.26Part 3 of 5(c) Compute the P-value. Round the answer to four decimal places.P-value=0.1038Part: 3 / 53 of 5 Parts CompletePart 4 of 5(d) Determine whether to reject H0. Use the =α0.05 level of significance.▼(Choose one) H0 at the =α0.05 level.