Introduction to Statistics
Homework #4: Lesson 6, Sections 5-7 through Lesson 7, Sections 1-4
Clearly Indicate Your Final Answer
Total Points Earned:
- ______ out of 6
- ______ out of 6
- ______ out of 6
- ______ out of 6
- ______ out of 4
- ______ out of 6
- ______ out of 3
- ______ out of 3
- ______ out of 4
- ______ out of 3
- ______ out of 3
Score ______ / 50
- (6 points) A sample of 100 plain M&M candies is randomly selected (without replacement) from a bag which contained a total of 465 M&M candies. The weight of each M&M (in grams) is recorded in the table below and in the available Excel Data Set file. From this data set, M&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0518 g. The package label for the M&M candy stated that the net weight of the candies contained within is 400 g.
Count | Red | Orange | Yellow | Brown | Blue | Green |
1 | 0.751 | 0.735 | 0.883 | 0.696 | 0.881 | 0.925 |
2 | 0.841 | 0.895 | 0.769 | 0.876 | 0.863 | 0.914 |
3 | 0.856 | 0.865 | 0.859 | 0.855 | 0.775 | 0.881 |
4 | 0.799 | 0.864 | 0.784 | 0.806 | 0.854 | 0.865 |
5 | 0.966 | 0.852 | 0.824 | 0.840 | 0.810 | 0.865 |
6 | 0.859 | 0.866 | 0.858 | 0.868 | 0.858 | 1.015 |
7 | 0.857 | 0.859 | 0.848 | 0.859 | 0.818 | 0.876 |
8 | 0.942 | 0.838 | 0.851 | 0.982 | 0.868 | 0.809 |
9 | 0.873 | 0.863 | 0.803 | 0.865 | ||
10 | 0.809 | 0.888 | 0.932 | 0.848 | ||
11 | 0.890 | 0.925 | 0.842 | 0.940 | ||
12 | 0.878 | 0.793 | 0.832 | 0.833 | ||
13 | 0.905 | 0.977 | 0.807 | 0.845 | ||
14 | 0.850 | 0.841 | 0.852 | |||
15 | 0.830 | 0.932 | 0.778 | |||
16 | 0.856 | 0.833 | 0.814 | |||
17 | 0.842 | 0.881 | 0.791 | |||
18 | 0.778 | 0.818 | 0.810 | |||
19 | 0.786 | 0.864 | 0.881 | |||
20 | 0.853 | 0.825 | ||||
21 | 0.864 | 0.855 | ||||
22 | 0.873 | 0.942 | ||||
23 | 0.880 | 0.825 | ||||
24 | 0.882 | 0.869 | ||||
25 | 0.931 | 0.912 | ||||
26 | 0.887 | |||||
27 | 0.886 |
- (1 point) If every package of M&M contains exactly 465 candies (they are counted by a machine before being placed in the bag) and the net weight is touted to be 400 g, what is the expected weight of each M&M candy?
- (2 points) If one M&M plain candy is randomly selected, find the probability that it weighs more than 0.8602 g.
- (2 points) If 465 M&M plain candies are randomly selected, find the probability that their mean weight is at least 0.8602 g.
- (1 point) Given these results, do you think the Mars Company puts exactly 465 M&M plain candies in a 400 g bag of plain M&Ms?
- (6 points) The known heights (in cm) of US presidents are shown in the table below:
President | Height (cm) | President | Height (cm) | |
Washington | 188 | McKinley | 170 | |
J. Adams | 170 | T. Roosevelt | 178 | |
Jefferson | 189 | Taft | 182 | |
Madison | 163 | Wilson | 180 | |
Monroe | 183 | Harding | 183 | |
J. Q. Adams | 171 | Coolidge | 178 | |
Jackson | 185 | Hoover | 182 | |
Van Buren | 168 | F. Roosevelt | 188 | |
Harrison | 173 | Truman | 175 | |
Polk | 173 | Eisenhower | 179 | |
Taylor | 173 | J. Kennedy | 183 | |
Pierce | 178 | Johnson | 192 | |
Buchanan | 183 | Nixon | 182 | |
Lincoln | 193 | Carter | 177 | |
Grant | 173 | Reagan | 185 | |
Hayes | 173 | G. H. W. Bush | 188 | |
Garfield | 183 | Clinton | 188 | |
Cleveland | 180 | G. W. Bush | 183 | |
Harrison | 168 | Obama | 188 |
- (3 points) Construct a histogram of the data and determine if the heights of US presidents appear to be normally distributed? Assume that this requirement is “loose” in the sense that the population distribution does not have to be exactly normal, but it must be a distribution that is roughly “bell-shaped.” NOTE: You might need to review lectures 3 & 4 to remember how to construct a histogram.
- (3 points) Generate a normal quantile plot for this data. What does this tell us about the data? NOTE: Excel will be extremely helpful for this problem, and the accompanying Excel Data File will keep you from having to type the data into Excel.
- (6 points) The cotinine level (in ng/mL) for nonsmokers who were environmentally exposed to tobacco smoke at home or work is listed in the table below. Note: cotinine is a metabolite of nicotine – when nicotine is absorbed into the body, cotinine is produced.
Subject # | Cotinine Level | Subject # | Cotinine Level | |
1 | 384 | 21 | 0 | |
2 | 0 | 22 | 3 | |
3 | 69 | 23 | 1 | |
4 | 19 | 24 | 45 | |
5 | 1 | 25 | 13 | |
6 | 0 | 26 | 3 | |
7 | 178 | 27 | 1 | |
8 | 2 | 28 | 1 | |
9 | 13 | 29 | 1 | |
10 | 1 | 30 | 0 | |
11 | 4 | 31 | 0 | |
12 | 0 | 32 | 551 | |
13 | 543 | 33 | 2 | |
-14 | 17 | 34 | 1 | |
15 | 1 | 35 | 1 | |
16 | 0 | 36 | 1 | |
17 | 51 | 37 | 0 | |
18 | 0 | 38 | 74 | |
19 | 197 | 39 | 1 | |
20 | 3 | 40 | 241 |
- (3 points) Construct a histogram and determine if the cotinine levels of people environmentally exposed to second hand smoke appear to be normally distributed? Assume that this requirement is “loose” in the sense that the population distribution does not have to be exactly normal, but it must be a distribution that is roughly “bell-shaped.”. NOTE: You might need to review lectures 3 & 4 to remember how to construct a histogram.
- (3 points) Generate a normal quantile plot for this data. NOTE: Excel will be extremely helpful for this problem, and the accompanying Excel Data File will keep you from having to type the data into Excel.
- (6 points) In 2010, there were 611 challenges made to referee calls in professional tennis singles play. Among those challenges, 172 were upheld with the call overturned. Assume that 30% of the challenges are successfully upheld with the call overturned.
- (1 point) Find the probability that among 611 challenges, the number of overturned calls is exactly 172.
- (2 points) Of the 611 challenges, the number of 172 overturned calls is fewer than 30%. Find the probability that among the 611 challenges, the number of overturned calls is 172 or fewer. If the 30% rate is correct, is 172 overturned calls for 611 challenges an unusually low number?
- (2 points) Which result is useful for determining whether the 30% rate is correct: part (a) or part (b)? Why?
- (1 point) Is there strong evidence to suggest that the rate of overturned calls is not 30%?
- (4 points) MicroSort’s XSORT gender-selection technique is designed to increase the likelihood that a baby will be a girl given in vitro fertilization was used. In updated resorts of the XSORT gender-selection techniques, 945 births resulted in 879 baby girls and 66 baby boys. In analyzing the results, assume that the XSORT method has no effect so that boys and girls are equally likely.
- (1 point) Find the probability of getting exactly 879 girls in 945 births.
- (1 point) Find the probability of getting 879 or more girls in 945 births. If boys and girls are equally likely, are 879 girls in 945 births unusually high?
- (1 point) Which probability is relevant for trying to determine whether the XSORT method is effective: the result from part (a) or the result from part (b)? Why?
- (1 point) Based on the results, does it appear that the XSORT method is effective? Why or why not?
- (6 points) [Note: This question revisits the data set from Problem 1] A sample of 100 plain M&M candies is randomly selected (without replacement) from a bag which contained a total of 465 M&M candies. The weight of each M&M (in grams) is recorded in the table below and in the available Excel Data Set file. From this data set, M&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0518 g. The package label for the M&M candy stated that the net weight of the candies contained within is 400 g.
Count | Red | Orange | Yellow | Brown | Blue | Green |
1 | 0.751 | 0.735 | 0.883 | 0.696 | 0.881 | 0.925 |
2 | 0.841 | 0.895 | 0.769 | 0.876 | 0.863 | 0.914 |
3 | 0.856 | 0.865 | 0.859 | 0.855 | 0.775 | 0.881 |
4 | 0.799 | 0.864 | 0.784 | 0.806 | 0.854 | 0.865 |
5 | 0.966 | 0.852 | 0.824 | 0.840 | 0.810 | 0.865 |
6 | 0.859 | 0.866 | 0.858 | 0.868 | 0.858 | 1.015 |
7 | 0.857 | 0.859 | 0.848 | 0.859 | 0.818 | 0.876 |
8 | 0.942 | 0.838 | 0.851 | 0.982 | 0.868 | 0.809 |
9 | 0.873 | 0.863 | 0.803 | 0.865 | ||
10 | 0.809 | 0.888 | 0.932 | 0.848 | ||
11 | 0.890 | 0.925 | 0.842 | 0.940 | ||
12 | 0.878 | 0.793 | 0.832 | 0.833 | ||
13 | 0.905 | 0.977 | 0.807 | 0.845 | ||
14 | 0.850 | 0.841 | 0.852 | |||
15 | 0.830 | 0.932 | 0.778 | |||
16 | 0.856 | 0.833 | 0.814 | |||
17 | 0.842 | 0.881 | 0.791 | |||
18 | 0.778 | 0.818 | 0.810 | |||
19 | 0.786 | 0.864 | 0.881 | |||
20 | 0.853 | 0.825 | ||||
21 | 0.864 | 0.855 | ||||
22 | 0.873 | 0.942 | ||||
23 | 0.880 | 0.825 | ||||
24 | 0.882 | 0.869 | ||||
25 | 0.931 | 0.912 | ||||
26 | 0.887 | |||||
27 | 0.886 |
- (1 point) You are provided with a confidence interval of 0.165 < p < 0.335(which is based on the proportion of red, orange, yellow, and blue M&Ms in the data set). Express this confidence interval in the form of .
- (1 point) You are provided with a confidence interval of (which is based on the proportion of red, orange, yellow, and blue M&Ms in the data set). Express this confidence interval in the form of .
- (3 points) The Mars candy company claims that 13% of its M&M candies are brown, but the 100 M&M candies sampled in the data set above, 8% of them are brown. Use the sample data to construct a 98% confidence interval estimate of the proportion of brown M&M candies.
- (1 point) Given your answer from part (c), does it appear that the claimed rate of 13% is wrong? Why or why not?
- (3 points) You have been given the task of estimating the percentage of Southwest flights that arrive on time, i.e. those flights that are no later than 15 minutes after the scheduled arrival time. How many flights must you survey in order to be 80% confident that your estimate is within three percentage points of the true population percentage?
- (2 points) Assume that nothing is known about the percentage of on-time Southwest flights.
- (1 point) Assume that for the recent year, 84% of Southwest flights were on time (based on data from the Bureau of Transportation)
- (3 points) A simple random sample of size n = 30was obtained from the population of duration times (in seconds) of eruption of the Old Faithful geyser in Yellowstone National Park?
- (1 point) What is the number of degrees of freedom that should be used for finding the critical value, ?
- (1 point) Find the critical value, , corresponding to n = 30and a 95% confidence interval.
- (1 point) Briefly describe what is meant by “number of degrees of freedom.”
- (4 points) The data set below contains the number of chocolate chips in a sample of 40 Chips Ahoy cookies. The mean of the sample is 23.95 chocolate chips, and the standard deviation is 2.55 chocolate chips.
22 | 22 | 26 | 24 | 23 | 27 | 25 | 20 | 24 | 26 |
25 | 25 | 19 | 24 | 20 | 22 | 24 | 25 | 25 | 20 |
23 | 30 | 26 | 20 | 25 | 28 | 19 | 26 | 26 | 23 |
25 | 23 | 23 | 23 | 22 | 26 | 27 | 23 | 28 | 24 |
- (2 points) Construct a 99% confidence interval estimate of the mean number of chocolate chips in all such cookies. How does the confidence interval not contradict the fact that most of the original values do not fall between the confidence interval limits?
- (2 points) Construct a 90% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.
- (3 points) You want to estimate the mean amount of time Internet users spend on Facebook each month.
- (2 point) How many Internet users must be surveyed in order to be 95% confident that your sample mean is within 15 minutes of the population mean? Based on results from a prior survey, assume that the standard deviation of the population of monthly times spent on Facebook is 210 minutes.
- (1 point) What is a major obstacle in getting a good estimate of the population mean?
- (3 points) 25 men and 25 women were assessed for health purposes, and their pulse rates recorded. The men’s pulse rates had a mean of 67.3 beats per minute and a standard deviation of 10.3 beats per minute. The women’s pulse rates had a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute.
- (1 point) Construct a 99% confidence interval estimate of the standard deviation of the pulse rates of men.
- (1 point) Construct a 99% confidence interval estimate of the standard deviation of the pulse rates of women.
- (1 point) Compare the variation of the pulse rats of men and women. Does there appear to be a difference?