In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend usingthe Internet in a typical week. The mean of the sample observations was 12.8 hours.(a) The sample standard deviation was not reported, but suppose that it was 4 hours. Carry out a hypothesis test with asignificance level of 0.05 to decide if there is convincing evidence that the mean time spent using the Internet byCanadians is greater than 12.6 hours. (Use a statistical computer package to calculate the P-value. Round your teststatistic to two decimal places and your P-value to four decimal places.)t =P-value =State the conclusion in the problem context.O Reject Ho. We do not have convincing evidence that the mean weekly time spent using the Internet byCanadians is greater than 12.6 hours.Do not reject Ho. We do not have convincing evidence that the mean weekly time spent using the Internet byCanadians is greater than 12.6 hours.O Reject Ho. We have convincing evidence that the mean weekly time spent using the Internet by Canadians isgreater than 12.6 hours.Do not reject Ho. We have convincing evidence that the mean weekly time spent using the Internet byCanadians is greater than 12.6 hours.(b) Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of0.05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greaterthan 12.6 hours. (Use a statistical computer package to calculate the P-value. Round your test statistic to twodecimal places and your P-value to four decimal places.)t =P-value =State the conclusion in the problem context.Reject Ho. We have convincing evidence that the mean weekly time spent using the Internet by Canadians isgreater than 12.6 hours.Reject Ho. We do not have convincing evidence that the mean weekly time spent using the Internet byCanadians is greater than 12.6 hours.Do not reject Ho. We have convincing evidence that the mean weekly time spent using the Internet byCanadians is greater than 12.6 hours.O Do not reject Ho. We do not have convincing evidence that the mean weekly time spent using the Internet byCanadians is greater than 12.6 hours.

It is given that Population mean = 12.6 Sample mean = 12.8 Sample Standard deviation = 4 Sample…

Answered: (a) The sample standard deviation was…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Consider a value to be significantly low if its z score less than or equal to −2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.4 and the standard deviation was 5.4. Identify the test scores that are significantly low or significantly high. What test scores are significantly low? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Test scores that are between ____ and _____. (Round to one decimal place as needed. Use ascending order.) B. Test scores that are less than _____. (Round to one decimal place as needed.) C. Test scores that are greater than _____. (Round to one decimal place as needed.)
The distribution of heights in a population of women is approximately normal. Sixteen percent of the women have heights less than 62 inches. About 97.5% of the women have heights less than 71 inches. Use the empirical rule to estimate the mean and standard deviation of the heights in this population. Mean: K inches Standard Deviation: inches
Use z scores to compare the given values. The tallest living man at one time had a height of 238 cm. The shortest living man at that time had a height of 142.4 cm. Heights of men at that time had a mean of 175.45 cm and a standard deviation of 5.59 cm. Which of these two men had the height that was more extreme? ... Since the z score for the tallest man is z = 0 and the z score for the shortest man is z = the man had the height that was Im- more extreme. (Round to two decimal places.) shortest tallest
Consider a value to be significantly low if its z score less than or equal to −2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 20.7 and the standard deviation was 4.7. Identify the test scores that are significantly low or significantly high. What test scores are significantly low? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Test scores that are between nothing and nothing. (Round to one decimal place as needed. Use ascending order.) B. Test scores that are greater than nothing. (Round to one decimal place as needed.) C. Test scores that are less than nothing. (Round to one decimal place as needed.) What test scores are significantly high? Select the correct answer below and fill in the answer box(es) to complete your choice. A. TEST SCORES THAT ARE GREATER…
After the pandemic hit and people were forced to spend more time indoors, there have been a suspicion among health professionals. that people are eating more and thus gaining weight. In 2018, the average weight of 192 pounds, and health experts want to know if that average weight of the American male has gone up during this pandemic. d. A sample of 50 American males yeilded a sample average weight of 195 pounds and a sample standard deviation of 14.5 pounds. What is your conclusion at the 5% level of significance? (be sure to state your conclusion in the context of this problem) e. Would your conclusion change if the level of significance were changed to 1%? Briefly explain.
Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.9. Assume that in a sample of 37 teenagers, the sample standard deviation of daily TV time is 2.2 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 2.9? Use the a=0.10 level of significance.
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
Microsoft Word -... # The average number of accidents at controlled intersections per year is 5.4. Is this average more for intersections with cameras installed? The 47 randomly observed intersections with cameras installed had an average of 5.9 accidents per year and the standard deviation was 1.08. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: V = H₁: > 5.4 c. The test statistic t = 3.174 (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? a f. Based on this, we should g. Thus, the final conclusion is that ... 80 F3 V 5.4 Select an answer the null hypothesis. O The data suggest that the populaton mean is significantly more than 5.4 at a = 0.01, so there is statistically significant evidence to conclude that the population mean number of accidents per year at intersections with…
Please answer B, C, & D.

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