Stat 4220 Homework
Due October 12
For hypothesis questions use all the steps of a hypothesis
1) A metal beam is supposed to support 1000 pounds. Assume the standard deviation for failure
times is 52 pounds. To test if these beams work you will randomly select 42 beams and measure
the weight they can hold before failing. Your average is 990 pounds. Does the data show with
α=0.05 that the beams are not working as specified?
H0: mu >= 1000
HA: mu < 1000
Alpha=0.05
Z=(990-1000)/(52/sqrt(42))=-1.25
p-value = 0.1056
Fail to reject
Our data does not show the beams are NOT working as specified
2) The amount of money the Emca company spends on tech support each month is random with a
standard deviation of $518. The company wants to know if they are spending $250 on monthly
average for tech support. They randomly select 12 months and get a sample average of $1,996.
What would their conclusion be using α=0.01?
Trick question – can’t do this one, it isn’t normal.
3) The amount of time it takes a random computer to load SuperVortex is normally distributed
with a standard deviation of 13 minutes. The pop-up window says it takes on average 30
minutes. I randomly selected 7 computers and loaded SuperVortex. My average was 42
minutes. Do I have evidence to disprove what the pop-up window says?
Ho: mu=30
Ha: mu ≠ 30
Alpha=0.05 (or something reasonable)
Z=(42-30)/(13/sqrt(7)) = 2.44
p-value = 2(.0073) = .0146
reject the null (unless they used alpha = 0.01)
Our data shows the average is not 30 minutes
4) A microwave will naturally exude a random amount of radiation. If the radiation level is 10 mSv
or above on average, then it can potentially cause cancer. Researchers are testing if a
microwave is safe before sending out on the market
a. What is the null and alternative hypothesis for their test?
H0: mu>=10 Ha:mu<10
b. What would be a type I error for this scenario?
Declaring it safe, when in fact it is not safe
c. What would be a type II error for this scenario?
Declaring it to be unsafe, when in fact it is fine
d. What would be the definition of power for this scenario?
How likely it is to know the microwave is safe if it really is
5) Ejection seats in the F150 jet push the pilot out of the cockpit with a random amount of force. If
the force is too low then the pilot may not be thrown a safe distance from the jet. If the force is
too high then the pilot could be injured from the ejection. The government has hired you to test
the new ejection seat. If you declare the seat unsafe it will cost the government 10 billion
dollars to redraw the specifications and improve the seat. You have been ordered to not use
α=0.05. What level of alpha would you pick, and why?
Two possible answers:
Small alpha because you don’t want the government spending 10 billion dollars
Large alpha because you don’t want pilots to die
6) The US government is spying on Switzerland. They believe Switzerland is developing a computer
virus which will melt all the computers in the US. They are sending in agent 00111 to investigate. He
will gather evidence, and if it is strong enough the president will have no choice but to declare war.
a) What is the definition of power in this scenario?
The null is that they do have a virus. Power is that they can tell when Switzerland is actually harmless
(think Iraq)
b) What alpha would you suggest and why (explain which consequence you are avoiding)
high alpha so that we don’t attack a country for no reason
or
low alpha so we can trounce on anyone who thinks about messing with the US
7) The March 2013 issue of Consumer Reports examined bagless upright vacuums and found an average of 18.4 lbs
with a standard deviation of 3.19 lbs. ”Vacuums” Consumer Reports March 2012 pg 44.
Assume that the weights normally distributed. The vacuum company NatureAbhors claims their vacuums are on
average lighter than the market average. We sample 5 bagless upright vacuum cleaners from Nature Abhors, and
get an average of 18.36. Can we show at the 5% significance level that their average is below the 18.4?
a. H0: mu >= 18.4
b. HA: mu < 18.4
c. α = 0.05
d. Z = -0.028
e. P-value = 0.488
f.
Fail to refect the null hypothesis
g.
Thus our data does not show their average is significantly below the national average
8) The Center for Disease Control reports that from 1999 - 2011 in the United States the number of
deaths related to exposure to excessive natural cold has an average of 1301 deaths per year with a
standard deviation of 129 deaths. Centers for Disease Control and Prevention. [Number of hypothermia- Related Deaths by
Sex]. MMWR 2013;61:[pg 18]. http://www.cdc.gov/mmwr/pdf/wk/mm6151.pdf pg 18
If we sample the next 36 years and get an average of 1260 deaths, will have enough evidence to say
the death rate from hypothermia is getting lower?
a. H0: mu>=1301
b. HA: mu<1301
c. α = whatever they pick
d. Z = -1.9
e. P-value = 0.0287
f.
At 0.05 reject the null hypothesis
g.
Thus our data would show the average is getting lower
9) A mechanical engineer builds a machine that makes mirrors consistently with a variance of 0.01
inches in width. The machine is supposed to be calibrated to make a mirror that is 0.05 inches in width,
but the QC manager says that it’s not making them thin enough. To find out the mechanical engineer
uses it to make 36 mirrors, and the average width from his sample was 0.046.
Use all 7 steps of a hypothesis test to determine who is right.
Notice that the variance is a design of the machine, and is assumed known without taking a sample to
estimate it. That means we know σ, so a Z-test is appropriate.
H0: μ ≤ 0.5
Ha: μ > 0.5 (this means “not thin enough”)
α = 0.05 (I just chose this)
Z
0.46  0.5
 2.4
0.01
36
Now if you don’t draw a picture you’re going to mess this up.
When you look up -2.4 on the Z-table you’ll get 0.0082, but you want the right tail (because the
alternative is “greater than”), so the pvalue is 1-0.0082
p-value = 0.9918
Fail to reject the null hypothesis
The data does not support the claim that the machine is not making the mirrors thin enough
10) The National Assessment of Educational Progress (NAEP) in science for its fourth grade
students in public schools assessment was updated in 2009 to keep pace with key developments
in science. Student performance is presented in terms of average scores on a scale from 0 to 300
with a mean of 150 and a standard deviation of 35. Scores are reported by state. The sample of
104 Wyoming students scored an average of 156. Can you prove that Wyoming students are
smarter on average than the rest of the nation?
http://www.nsf.gov/statistics/seind12/pdf/c08.pdf pg. 16
H0: mu <= 150
HA: mu > 150
Alpha = 0.05 (my pick)
Z=(156-150)/(35/sqrt(104)) = 1.75
p-value = 0.0401
Reject the null
We can show the Wyoming students are smarter on average than the rest of the nation.
11) An automatic cement mixer dumps in one 50 kg bag of cement mix and then sprays in 22 liters of
water. Unfortunately the city water pressure fluctuates, meaning you don’t get exactly 22 liters each
time. The good news is that the huge cement mixer repeats this process 120 times to fill the cement
truck. If the average for each of the 120 sprays is 22 liters, then the cement will mix well. Assume each
time the mixer sprays water is independent (random) of the previous time, and that σ=0.6 liters.
To test if the mixer is working properly an employee of the Sretsbom Company sends an engineer to
measure the exact amount of water from 120 sprays. He gets an average of 22.1 liters. Can he say at
the 5% significance level the mixer is off on average?
H0: mu=22
HA: mu≠22
Alpha=0.05
Z=(22.1-22)/(.6/sqrt(120)) = 1.83
P-value = 2*0.0336=0.0672
Fail to Reject
We cannot say the mixer is off target on average.