At a large university in Conway, South Carolina, the Dean wants to know what proportion of students enjoyed their introductory statistics course. In order to answer this question, 200 randomly selected students who have taken an introductory statistics course were contacted and asked whether they enjoyed this course. 145 out of the 200 answered that they enjoyed the course. Using this data, construct a 99% confidence interval for the true proportion of students who enjoyed their introductory statistics course. (0.6631,0.7869) (0.7206,0.7294) (0.6437,0.8063) (0.6731,0.7769)

Previously, an organization reported that teenagers spend 8.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Twenty-three randomly chosen teenagers were asked how many hours per week they spend on the phone, with the following results (in hours): 8.3, 8.5, 10.5, 2.9, 6.7, 7.7, 13.8, 7.8, 8.7, 11.8, 12, 6.1, 5.4, 5.8, 8.5, 7.9, 10.1,11.1, 11.6, 12.9, 3.6, 6.4, 11.2 Perform a hypothesis test using a 3% level of significance. Step 1: State the null and alternative hypotheses. Ho: ??v H: ? v ? v (So we will be performing a Select an answer v test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. v with By the Central Limit Theorem, we know that the point estimates are Select an answer and distribution standard deviation distribution mean Step 3: Find the p-value of the point estimate. P? v? v = P(| ? ♥| ? v

A Pew Internet Project Data Memo presented data comparing adult gamers with teen gamers with respect to the devices on which they play. The data are from two surveys. The adult survey had 1063 gamers, while the teen survey had 1064 gamers. The memo reports that 54% of adult gamers played on game consoles (Xbox, PlayStation, Wii, etc.), while 89% of teen gamers played on game consoles.  Suppose you want to construct a 95% confidence interval for the difference.  You want to test were or not the two proportions (adult using consoles vs teens using consoles) are the same.  You have to calculate a pooled estimate for the proportion.  What would that be?  (Please round to three decimal places)

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Construct a 95% confidence interval for the true mean or proportion. Round to the nearest hundredth (2 decimal places). (  ,  )

From candy to jewelry to flowers, the average consumer was expected to spend $104.63 for Mother’s Day in 2005, according to a Democrat & Chronicle article. Local merchants thought this average was too high for their area.  They contracted an agency to conduct a study. A random sample of 60 consumers was taken at a local shopping mall the Saturday before Mother’s Day and produced the sample mean amount of $94.27.  If σ = $29.50, does the sample provide sufficient evidence to support the merchants’ claim at the 0.03 level of significance? 1. Do you recognize any flaws in this study and what conclusions can you draw?     The article “Well-fed Crickets Bowl Maidens Over” reports that female field crickets are attracted to males that have high chirp rates and hypothesized that chirp rate is related to nutritional status. The usual chirp rate doe male field crickets was reported to vary around a mean of 60 chirps per second. To investigate whether chirp rate was related to nutritional…

According to recent data by the American Marketing​ Association, 76.5​%of people in the​ 18-24 age group used social media in 2017. You want to find out what the social media proportion is among your classmates who are all between 18 and 24 years old. Suppose you obtain a random sample of 229 students and find that 202 students in this sample are active social media users. Complete parts a through c below.   a. Construct a 99​% confidence interval to estimate the actual proportion of social media users. A 99​% confidence interval to estimate the actual proportion has a lower limit of------ and an upper limit of------- ​(Round to three decimal places as​ needed.) b. What is the margin of error for this​ sample? c. Is there evidence that this proportion matches that of the overall population based on the​ sample?

A statistics content developer at Aplia wanted to know whether study skills are related to memory quality. She invited student volunteers to perform an online memory task. The students saw a list of 60 words and were then asked to recognize a list of 10 words that were on the original list. Students were also asked to provide their GPAs.   Consider the following data set, which was collected from student volunteers in 2009. The table gives the frequency for five intervals of scores on the number of correctly identified words. Use the dropdown menus to complete the table by filling in the missing values for the proportions and percentages. Score Interval f Proportion Percentage 9–10 29 0.19 19% 7–8 53           5–6 50           3–4 22 0.14 14% 1–2 1 0.01 1%

The U.S. National Center for Health Statistics complies birth statistics and publishes the results in vital statistics of the United States. The Center has historically believed that the mean of all newborns in the U.S. has been no less than 6.95 pounds. A sample of babies is selected so that the center can revise its belief. The following birth weights, in pounds are recorded: 7.4, 6.0, 8.6, 4.5, 2.0, 7.9, 4.0, 2.6, 5.9, 7.3, 7.3, 7.0, 6.3, 8.1, 7.1, 7.3, 6.6, 5.2, 9.8, 8.0, 10.9, 6.5, 3.8, 5.0, 8.0 a). At 5% level of significance, do these data provide sufficient evidence to refute the Center’s belief? Your conclusion must be in terms of the P-Value as well as setting up a Rejection Region. You must show all necessary work. b). Which statistical distribution should be applied in this situation and why? Explain carefully.

A psychologist wanted to know if students in her class were more likely to cheat if they were low achievers. She divided her 60 students into three groups (low, middle, and high) based on their mean exam score on the previous three tests. She then asked them to rate how likely they were to cheat on an exam if the opportunity presented itself with a very limited chance for consequences. The students rated their desire to cheat on a scale ranging from 1-100, with lower numbers indicating less desire to cheat. The data: Achievement_Group Cheat 1 20 1 40 1 49 1 50 1 51 1 51 1 52 1 53 1 58 1 42 1 48 1 48 1 52 1 55 1 55 1 56 1 59 1 67 1 80 1 79 2 19 2 25 2 20 2 29 2 24 2 32 2 25 2 27 2 30 2 55 2 40 2 25 2 27 2 35 2 42 2 30 2 30 2 34 2 40 2 27 3 60 3 65 3 69 3 78 3 79 3 80 3 80 3 90 3 95 3 50 3 55 3 55 3 60 3 69 3 70 3 70 3 88 3 90 3 90 3 91   Before opening the data,…