Stat 4220 homework – Due April 24
1)The casino boss has heard a rumor that the dice in his casino are rigged. (For those not familiar with dice, there are 6 possible outcomes, and each outcome is supposed to be equally likely). He hired a graduate student (lackey) to take a die and roll it 1002 times. He recorded that the die rolled a one 150 times, rolled a two 155 times, rolled a three 164 times, rolled a four 180 times, rolled a five 150 times, and rolled a six 203 times. Test whether the die is rigged.
H0: The dice are not rigged (all proportions are 1/6)
Ha: The dice are rigged (not all proportions are 1/6)
α=0.05
1 / 2 / 3 / 4 / 5 / 6167 / 167 / 167 / 167 / 167 / 167
1 / 2 / 3 / 4 / 5 / 6
150 / 155 / 164 / 180 / 150 / 203
1 / 2 / 3 / 4 / 5 / 6
1.73 / .86 / .054 / 1.012 / 1.73 / 7.76
Chi-squared = 13.15
.02<p-value<.025
Reject
Our data shows the dice is rigged
2)The same group of students also indicated on the survey what year in school they were. Since statistics is a sophomore level class, we might expect to see more sophomores than any other group. Use a chi-square goodness-of-fit test to determine if the different years in school are equally distributed.
Year / Freshman / Sophomore / Junior / Senior50 / 65 / 37 / 20
H0: The distribution across the classes is even
Ha: The distribution across the classes is uneven
α=0.05
Year / Freshman / Sophomore / Junior / Senior43 / 43 / 43 / 43
Year / Freshman / Sophomore / Junior / Senior
1.14 / 11.26 / 0.84 / 12.30
Χ2=25.54
p-value < 0.0001
Reject
The distribution of classes is not even
3)For many years TV executives used the guideline that 30% of the viewing audience were watching each of the traditional big three prime-time networks and 10% were watching cable stations on a weekday night. A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder were viewing a cable station. At the .05 significance level, can we conclude that the guideline is still reasonable?
H0: The distribution given is correct
HA: The distribution has changed
Alpha=0.05
Observed: 165,140,125,70
Expected: 150,150,150,50
Chisquar: 1.5, .67, 4.16, 8
=14.33
p-value = 0.002
Reject
The distribution has changed
4)A 98% confidence interval for the difference in the average length of a movie between “action flicks” and “chick flicks” was (22.43, 25.61) minutes. Which of the following statements is true?
98% of “action flicks” and “chick flicks” last between 22.43 minutes and 25.61 minutesThe probability the next “action flick” or “chick flick” lasts between 22.43 and 25.61 minutes is 98%
We are 98% confident the average time for “chick flicks” and “action flicks” is between 22.43 and 25.61
The evidence does not support the claim that “action flicks” have a different average than “chick flicks”
X / “Action flicks” are 22.43 to 25.61 longer than “chick flicks” on average with 98% confidence
Of all possible “action flicks” and “chick flicks” 98% have an average difference of 22.43 to 25.61
5)University of Michigan surveyed high school seniors nationwide who smoke and asked them which brands of cigarettes they use. Is there a relationship between Race and Cigarette Brand?
LD Johnston, PM O'Malley, JG Bachman, JE Schulenberg. (Apr. 1999). Cigarette brands smoked by American teens: One brand predominates; three account for nearly all of teen smoking. University of Michigan News and Information Services: Ann Arbor, MI. [On-line]. Available: accessed 04/15/2013
Black / White / HispanicMarlboro / 6 / 1276 / 90
Newport / 87 / 138 / 36
Camel / 0 / 198 / 5
All other Brands / 13 / 205 / 25
H0: race is independent of cigarette preference
Ha: cigarette preference depends on race
Alpha:0.05
Black / White / HispanicMarlboro / 69.95286 / 1199.098 / 102.9495
Newport / 13.30736 / 228.1082 / 19.58442
Camel / 10.35017 / 177.4175 / 15.23232
All other Brands / 12.38961 / 212.3766 / 18.23377
Black / White / Hispanic
Marlboro / 58.46749 / 4.932019 / 1.628851
Newport / 408.0904 / 35.59491 / 13.75948
Camel / 10.35017 / 2.387808 / 6.87357
All other Brands / 0.030072 / 0.256217 / 2.510832
Chisq:544.88
p-value = 0
Reject
Certain cigarettes do depend on race
6)A study investigated whether people think Labrador retrievers are cuter than Afghan Hounds. They walked a Labrador past 100 people and 78 petted the dog. They walked an Afghan Hound past 90 people and 61 petted the dog. Find a 96% confidence interval for the difference in proportions of people who will pet a Labrador verses an Afghan Hound.
78/100-61/90+-2.054*sqrt(78/100*(1-78/100)/100+61/90*(1-61/90)/90)=(-0.02972, 0.23416)
7)George Bush Sr. mentions on T.V. that the average age of a student at UW is 23 years old. To test his hypothesis, you ask 3 randomly chosen UW students what their ages are, and use α=.01 Assume the ages of students at A&M are normally distributed.
The ages were : 22 years old, 28 years old, and 24 years old.
Test whether George Bush was right.
n = 3
= (22+28+24)/3 = 24.67
μ0 = 23
df = 3-1 = 2
sx =
H0: μ = 23
HA: μ ≠ 23
α=.01
The ages are normally distributed, so we can use a t-test
The t-value of .94 is between .817 and 1.061 on the t-table for 2 degrees of freedom, so the tail-probability is between .25 and .2. This is a two-sided test, so we need to double the probability to get the p-value.
.4 < p-value < .5
Since .25 > .01 we cannot reject the null hypothesis.
So Bush was right. (Again), it appears that the average age of students at UW is about 23 years old.
8)The Working Imitation Design Gadget Engineering Tool is manufactured by a machine that sometimes has a flaw in the production. According the machine specs the flaw distribution per hour should be:
0 flaws / 1 flaw / 2 flaws / 3 flaws / 4 flaws / 5 flaws70% / 15% / 8% / 3% / 2% / 1%
To see whether the machine is performing according to specifications we randomly sample 200 hours and get the following flaw distribution:
0 flaws / 1 flaw / 2 flaws / 3 flaws / 4 flaws / 5 flaws131 / 28 / 16 / 11 / 8 / 6
Can we say with 5% significance that the machine is not performing at specifications?
There is not enough data to do a goodness of fit test – an expected value is less than 5
9)Billy says that the average speed of a mule is faster than the average speed of a zebra. To test this he rides 7 different zebras and 7 different mules. Assume each time he rides the animal the exact same way. Test whether Billy is right.
Zebra / MuleFirst ride / 31 / 42
Second ride / 22 / 37
Third ride / 40 / 28
Fourth ride / 28 / 39
Fifth ride / 35 / 31
Sixth ride / 37 / 48
Seventh ride / 34 / 33
MEAN / 32.43 / 36.85
S / 6.02 / 6.87
H0:mu1<=mu2
Ha: mu1>mu2
Alpha: 0.05
T=(32.43-36.85)/(6.02^2/7+6.87^2/7) = -1.28
p-value = .12
Fail to Reject
We cannot say the mules run faster than the zebras
10)In a class survey done in a statistics class, students were asked, “Suppose that you are buying a new car and the model you are buying is available in three colors: silver, blue, or green. Which color would you pick?” Of the 111 students who responded, 59 picked silver, 27 picked green, and 25 picked blue. Is there sufficient evidence to conclude that the colors are not equally preferred?
H0: colors equally preferred
Ha: colors not equally preferred
Alpha=0.05
O: 59,27,25
E: 37,37,37
X2: 13.08,2.7,3.89
X2=19.68
p-value = 0
Reject
Colors not equally preferred
11)Suppose that on a typical day, the proportion of students who drive to campus is .30 (30%), the proportion of students who bike is .60 (60%), and the remaining .10 (10%) come to campus (e.g., walk, take the bus, get a ride). The campus sponsors a “spare the air” day to encourage people not to drive to campus on that day. They want to know whether the proportion using each mode of transportation on that day differ from the norm. To test this hypothesis, a random sample of 300 students that day was asked how they got to campus, with the following results:
Method of Transportation / Drive / Bike / Other / TotalFrequency / 80 / 200 / 20 / 300
H0:New program did nothing
Ha: New program effected a change
Alpha=0.05
O:80,200,20
E:90,180,30
X2:1.11,2.22,3.33
X2:6.667
p-value 0.035
Reject
The program did effect change
12)Ashley is eating ice cream when she gets a brain freeze. Her thought is that it’s because she was eating with her left hand. So she gets 100 bowls of cookie dough ice cream and asks 50 of her friends to randomly choose either their right or their left hand. They eat as fast as they can until they get a brain freeze. Then she asks them to switch hands with a new bowl of cookie dough ice cream and eat until they get a brain freeze. Here is her data:
Left hand:
50 bowls
Average time: 48 seconds
Standard Deviation: 27 seconds
Right hand:
50 bowls
Average time: 37 seconds
Standard Deviation: 24 seconds
Pooled Standard deviation: 25.5 seconds
Matched Pairs deviation: 2.5 seconds
Difference in deviations: 3 seconds
Make a 99% Confidence interval for the difference in times for each hand
It doesn’t say it’s normal, but n=50, so the mean is normal. This is matched pairs because each right hand is paired to one of the left hands by being attached to the same body.
t score= 2.704
(48-37) ± 2.704 * 2.5 / sqrt(50)
(10.04, 11.95)
13)The MagBlastcompany demolishes buildings by setting four charges on each corner of the building. The charges are supposed to detonate at the same time, but sometimes something goes wrong and not all of them ignite on time. If you had 500 buildings the distribution for the expected number of charges that would go off on time is given below:
All five detonate / Only four / Only three / Only two / Just one detonates / No charges detonate406 / 35 / 27 / 15 / 10 / 7
You have been asked to investigate what would happen to the charges if a building was demolited when it is raining. To find out you randomly select 500 buildings around the United States and demolish them when it is raining. The data you observed is given below:
All five detonate / Only four / Only three / Only two / Just one detonates / No charges detonate417 / 25 / 31 / 17 / 10 / 0
Test whether your data supports the hypothesis that rain affects the detonation of the charges (assuming homeland security does not catch you).
H0: The rain does not affect the detonation (proportions remain the same)
Ha: The rain does affect the detonation (at least one proportion is different)
Alpha = 0.05
All five detonate / Only four / Only three / Only two / Just one detonates / No charges detonate.298 / 2.857 / .5926 / .2667 / 0 / 7
Chisq=11.01
.05<p-value<.06
Fail to Reject
Our data does not show that rain affects the detonation
14)Captain Buckwheat uses a sextant to measure the height of a ship when he spots it on the horizon. After attacking the ship he calculates the gold looted. His goal is to be able to predict the amount of gold based on the ship height. Below is the data and regression from the 500 ships he has attacked during his career.
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.8958 0.1879 52.66 <2e-16 ***
Height 6.5537 0.3154 20.78 <2e-16 ***
Based on the output above, are there any assumptions that you feel should be investigated?
Perhaps there is clumping, which might indicate a nonrandom sample
Based on the output above find a 90% confidence interval for the slope of the regression line.
6.5537+-1.65*0.3154=(6.03, 7.07)
15)Every day I see the elevator says it’s been “inspected.” Somehow I feel dubious. I think the elevator gets inspected less than 70% of the time. I plan on doing a hypothesis test by investigating 100 elevators and using α=0.10. If the true percentage was actually only 65%, how powerful would my test be?
Z=-1.28, critical value = 0.64127
0.427 is the best answer, but if they use the null proportion in the alternative distribution (which I think is a reasonable error, easy to make) they’ll get .424
16)Randomly selected deaths of motorcycle riders are summarized in the table below. Use a .05 significance level to test the claim that such fatalities occur with equal frequency in the different months.
Month / Jan / Feb / Mar / Apr / May / June / July / Aug / Sept / Oct / Nov / Dec / TotObserved / 6 / 8 / 10 / 16 / 22 / 28 / 24 / 28 / 26 / 14 / 10 / 8 / 200
Expected / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 16.7 / 200
X2 / 6.86 / 4.53 / 2.7 / 0.03 / 1.68 / 7.64 / 3.19 / 7.64 / 5.18 / 0.44 / 2.69 / 4.53 / 47.1
H0: Even across the months
Ha: Not even
Alpha:0.05
X2:47.1
p-value =0
reject
not even across the months