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.

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.Here are the null and alternative hypotheses for her study: H0: µ = 96, Ha: µ > 96.What does µ represent in these hypotheses? Mean number of boxes of cookies sold for the average Girl Scout in New York City Mean number of boxes of cookies sold for the Girl Scouts in her sample from Troop 5078 Mean number of boxes of cookies sold for the Girl Scouts in Troop 5078

Are you smarter than a second grader? A random sample of 70 second graders in a certain school district are given a standardized mathematics skills test. The sample mean score is =x55. Assume the standard deviation of test scores is =σ21. The nationwide average score on this test is 50. The school superintendent wants to know whether the second graders in her school district have greater math skills than the nationwide average. 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  =μ50:H1  >μ50This hypothesis test is a ▼right-tailed test.Part 2 of 5(b) Compute the value of the test statistic. Round the answer to two decimal places.z=1.9Correct Answer:=z1.99Part: 2 / 52 of 5 Parts CompletePart 3 of 5(c) Compute the P-value of the test statistic. Round the answer to four decimal places.P-value=Skip PartCheckSave For LaterSubmit Assignment

he restaurant manager is testing the bartender's ability to pour 45 mL of spirits correctly into a mixed drink. The manager has the bartender pour water into 12 shot glasses to test their ability to pour the correct amount of spirits: 48 45 44 43 46 47 42 46 47 45 47 49 Note: The data appears to be approximately normally distributed. Test the bartender's ability to pour 45 mL at the 5% level of significance. T-Distribution Table a. Calculate the sample mean and standard deviation. x̄ = 45.750 Round to three decimal places if necessary s= 0.000 Round to three decimal places if necessary b. Calculate the test statistic. t= 0.000 Round to three decimal places if necessary c. Determine the critical value(s) for the hypothesis test. -2.201 ×2.201 × + Round to three decimal places if necessary d. Conclude whether to reject the null hypothesis or not based on the test statistic. Reject Fail to Reject

We collect random samples of 25 students at a time and calculate the proportion of females in each sample. The standard deviation of p-hats is approximately 0.10. Which of the following is a plausible standard deviation for samples of 100?

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.