Question

When a survey​ asked, "About how many hours per week do you spend sending and answering​ e-mail?" the 7 females in the survey sample of age at least 80 had the following responses. Use the data to complete parts a through c. 0 0 3 4 6 8 13 a. Using​ technology, find the sample​ mean, standard​ deviation, and the standard error of the sample mean. The sample mean is 4.86. ​(Round to two decimal places as​ needed.) The sample standard deviation is 4.63. ​(Round to two decimal places as​ needed.) The standard error of the sample mean is 1.75. ​(Round to two decimal places as​ needed.) b. Using​ technology, find and interpret a 90​% confidence interval for the population mean. The 90​% confidence interval for the population mean is left parenthesis nothing comma nothing right parenthesis . ​(Round to one decimal place as​ needed.) Interpret the confidence interval from the previous step. Choose the correct answer below. A. There is 0.95​% confidence that the population mean for the number of hours per week people spend sending and answering​ e-mail is between these two values. B. There is 90​% confidence that the population mean for the number of hours per week people spend sending and answering​ e-mail is between these two values. Your answer is correct.C. There is 0.90​% confidence that the population mean for the number of hours per week people spend sending and answering​ e-mail is between these two values. D. Approximately 90​% of the population send and answer​ e-mail between these two values in hours per week. c. Explain why the population distribution may be skewed right. If this is the​ case, is the interval you obtained in part b​ useless, or is it still​ valid? Explain why the population distribution may be skewed right. Choose the correct answer below. A. Since there will be many women that are at least 80 years of age who do not use​ e-mail at all and nobody who uses​ e-mail frequently, the distribution is likely to be skewed right.   Your answer is not correct.B. Since there will be many women that are at least 80 years of age who do not use​ e-mail at all and many who use​ e-mail frequently, the distribution is likely to be skewed right.   C. Since there will be many women that are at least 80 years of age who do not use​ e-mail at all but some who use​ e-mail frequently, the distribution is likely to be skewed right.   This is the correct answer. If the population distribution is skewed​ right, is the interval you obtained in b​ useless, or is it still​ valid? Valid Your answer is correct. Useless

Answer 1

a) As already given, sample mean, = 4.86, sample std dev. = 4.73

Sample Std Error, SE = Sample Std Dev / = 1.79

b) 90% confidence interval for population mean lies in the middle 90% area of the normal curve, so we need to boundaries of this area in the std normal curve (the left and right 5% boundaries)

z_0.05 = 1.645

CI = +/- z_0.05 * SE = 486 +/- 1.645*1.79 = (1.915, 7.805)

By definition of confidence interval, there is 90% confidence that the above interval (1.915, 7.805) is the interval in which the population mean of the no. of hours spend sending and answering e-mail lies. Hence, correct option is B

c) The given frequency data is: 0 0 3 4 6 8 13, hence there could be many women who do not use emails at all while fewer using emails very frequently. Hence, the lower numbers will have a high probability in this case, while higher numbers will have a low probability. This indicates a longer tail towards the right, hence the population distribution could be right skewed. Hence, correct answer is C

d) The confidence interval around the sample mean is useful when the population distribution is symmetric around the mean. If it's skewed, a confidence interval around the median could make more sense. Hence the confidence interval constructed above becomes useless in this scenario.