Assignment 10

Ho: p = .25 Ha: p ≠ .25 test statistic = 4.38 p-value = 0 (See: https://imgur.com/lbyP8Df) Yes

A telephone survey of 1000 randomly selected US adults found that 31 % of them say they believe in ghosts.1 Does this provide evidence that more than 1 in 4 US adults believes in ghosts? Clearly state the null and alternative hypotheses. Calculate the test statistic and p-value. Round your answer for the test statistic to two decimal places, and your answer for the p-value to three decimal places. Can we conclude that more than 1 in 4 US adults believes in ghosts?

The 90 % confidence interval is .75 to 6.07

How big is the home field advantage in the National Football League (NFL)? To investigate this question, we select a sample of 80 games from the 2011 regular season and find the home team scored an average of 25.16 points with a standard deviation 10.14 points. In a separate sample of 80 different games, the away team scored an average of 21.75 points with a standard deviation of 10.33 points. Use this summary information to find a 90 % confidence interval for the mean home field advantage, μH minus mu μA, in points scored. Round your answers to two decimal places.

The 99% confidence interval is .799 to .861 .031 No Yes

In a nationwide poll of 1000 randomly sampled adults conducted in June 2011, 83 % said they think children spend too much time on their computers and other electronic devices (but 37 % say time spent on a computer is better than time spent in front of a TV). Find a 99% confidence interval for the proportion of adults who believe children spend too much time on electronic devices. Round your answers to three decimal places. What is the margin of error for this result? Round your answer to three decimal places. Is it plausible that the proportion of all adults who feel this way is less than 79 %? Is it plausible that the proportion is greater than 84 %?

The 99% confidence interval is 5.747 to 7.261

In the dataset StudentSurvey, 361 students recorded the number of hours of television they watched per week. The average is x̅ equals 6.504 hours with a standard deviation of 5.584. Find a 99% confidence interval for μ. Round your answers to three decimal places.

Ho: μ = 12 Ha: μ ≠ 12 test statistic = -1.685 p-value = .1 Conclusion: do not reject Ho (See: https://imgur.com/fpt5scI) No

Susan is in charge of quality control at a small fruit juice bottling plant. Each bottle produced is supposed to contain exactly 12 fluid ounces (fl oz) of juice. Susan decides to test this by randomly sampling 30 filled bottles and carefully measuring the amount of juice inside each. She will recalibrate the machinery if the average amount of juice per bottle differs from 12 fl oz at the 1 % significance level. The sample of 30 bottles has an average of 11.92 fl oz per bottle and a standard deviation of 0.26 fl oz. State the null and alternative hypotheses. Give the test statistic and the p-value and state the conclusion of the test. Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places. Should Susan recalibrate the machinery?

Ho: μ = 1 Ha: μ < 1 test statistic = -10.098 p-value = .001 Yes

The US Food and Drug Administration has a limit for mercury content in fish of 1.0 ppm (parts per million), while in Canada the limit is 0.5 ppm. Use the variable Avg_Mercury in the FloridaLakes dataset to test whether there is evidence that average mercury level of fish (large-mouth bass) in Florida lakes is less than 1.0 ppm and whether there is evidence that average mercury level of fish in Florida lakes is less than 0.5 ppm. (a) Less than 1.0 ppm. State the null and alternative hypotheses. Give the test statistic and the p-value. Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places. Does the sample provide evidence that the average mercury level of fish (large-mouth bass) in Florida lakes is less than 1.0 ppm?

Males watch more TV by 2.383 hours per week.

The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: (See: https://imgur.com/2pdv0j1) (a) In the sample, which group watches more TV, on average? By how much? Enter the exact answer.

The 99 % confidence interval is -3.83 to -.94

The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: (See: https://imgur.com/2pdv0j1) (b) Use the summary statistics to compute a 99 % confidence interval for the difference in mean number of hours spent exercising μf minus μm, where μf represents the number of hours spent watching TV per week by all female students at this university and μm represents the number of hours spent watching TV per week by all male students at this university. Round your answers to two decimal places.

Same up to round-off error

The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: (See: https://imgur.com/2pdv0j1) (c) Compare the answer from part (b) to the confidence interval given in the following computer output for the same data: (See: https://imgur.com/3Ead2DN)

point estimate = .653 margin of error = +/- .096 The 95% confidence interval is .557 to .748

Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 95% confidence interval for the proportion who will answer ''Yes" to a question, given that 62 answered yes in a random sample of 95 people Round your answers to three decimal places.

The 95% confidence interval is 5.565 to 6.835

Use the sample information below and the t-distribution to find a 95% confidence interval for the population mean μ. Assume the underlying distribution is approximately normal. n equals 30 x̅ equals 6.2 s equals 1.7 Round your answers to three decimal places.

point estimate = 11 margin of error = 1.643 The 95% confidence interval is 9.357 to 12.643

Use the t-distribution to find a confidence interval for a mean mu given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 95% confidence interval for μ using the sample results x̅ equals 11.0, s equals 4.4, and n equals 30. Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.

Ho: p = .1 Ha: p ≠ .1 test statistic = 1.79 p-value = .074 (See: https://imgur.com/BCYFF7l) 5%: do not reject 10%: reject

Approximately 10 % of Americans are left-handed (we will treat this as a known population parameter). A study on the relationship between handedness and profession found that in a random sample of 105 lawyers, 16 of them were left-handed. Test the hypothesis that the proportion of left-handed lawyers differs from the proportion of left-handed Americans. Clearly state the null and alternative hypotheses. Let be the proportion of left-handed lawyers. Calculate the test statistic and p-value. Round your answer for the test statistic to two decimal places, and your answer for the p-value to three decimal places. What do we conclude at the 5 % significance level? At the 10 % significance level?

Ho: μ = 634 Ha: μ ≠ 634 test statistic = 1.59 p-value = .11 Conclusion: do not reject Ho (See: https://imgur.com/fpt5scI) No

Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. It is nearly impossible for most people to reliably list all the people they know, but using a mathematical model, social analysts estimate that, on average, a US adult has social ties with 634 people. A survey of 1700 randomly selected US adults who are cell phone users finds that the average number of social ties for the cell phone users in the sample was 664 with a standard deviation of 778. Does the sample provide evidence that the average number of social ties for a cell phone user is significantly different from 634, the hypothesized number for all US adults? State the null and alternative hypotheses. Give the test statistic and the p-value and state the conclusion of the test. Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places. Does the sample provide evidence that the average number of social ties for a cell phone user is significantly different from 634, the hypothesized number for all US adults?