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1. Suppose the prices of new laptops are normally distributed with a mean of $1300 and a standard deviation of
a) What percentage of new laptops cost less than $900?
900  1300
 1.333333333
the table says .0918, so 9.18% of laptops cost less than $900.
b) What percentage cost between $1105 and $1730?
Find each z score individually:
1105  1300
 0.65 the table says 0.2578
1730  1300 430
 1.433333 the table says 0.9236
Since we want between these, we’ll subtract:
0.9236 – 0.2578 = 0.6658 so 66.58% of laptops are between $1105 and $1730
c) Kevin says his laptop cost more than 95% of all other laptops. How much did his laptop cost?
Start with the table. Table says .9505 when z = 1.65. Put this into the equation and solve for x.
1.65 
x  1300
1.65(300)  x  1300
1.65(300) 1300  x
So x = $1795.
2. I want to find out the mean length of all Radiohead songs, but I'm kind of lazy, so instead of actually finding the
information for all Radiohead songs, I just look through 23 songs from a few different cd's I have. These 23 songs
have a mean length of 3:28. It turns out, the actual mean length of all Radiohead songs is 3:52. For this
scenario, identify the following:
population _All Radiohead songs in the whole universe____ parameter value _3:52_____
sample __the 23 songs I sample___________________
statistic value __3:28_____
3. For each of the following statistics, state the population which is being described and the variable being
“16% of students pay for some form of training or coaching when preparing for the SAT”
Population – students preparing for the SAT
Variable – whether or not they pay for coaching
“Support for the new health care legislation is at 49%”
Population – adult Americans
Variable – whether or not they approve of health care legislation
“Caffeine has been shown to reduce muscle soreness after exercising by 18%”
Population – athletes
Variable – level of muscle soreness
4. A pharmaceutical company believes its new supplement improves focus and learning ability and claims it will
improve student performance.
Using the five step process described in class, give a detailed outline of an experiment designed to test this claim.
When appropriate, use specific terms such as: bias, confounding variable, control group, double-blind, placebo,
population, sample, single-blind, and others.
Step 1 Identify Population and Variable
The supplement could probably be taken by anyone, but this study is focused on students. They’re probably not
talking about my first grade daughter, so I think the population they have in mind is students ages 12 and up (or
so). The thing they’re trying to measure is their focus – presumably they’d have to test them over some
information they’re given in a potentially distracting setting. Peter Piper Pizza, for example.
Step 2 Get a Sample
Select a group of students. There will hopefully be variety among their ages and academic abilities.
Step 3 Divide Sample Into Groups
We want to try to test if this supplement improves focus, so we will put them all through some sort of information
in – information out scenario and some of them will be given the supplement beforehand and some won’t. We
could make the experiment single-blind by giving those who don’t receive the real supplement a sugar pill
(placebo). We could make it double-blind by hiring other people to conduct the quizzing (say, Peter Piper
employees) so that the observers would not know who had received the supplement.
Step 4 Impose Treatment
I’m envisioning something like a huge screen with images passing across it while lights are flashing and bells and
whistles are going off – seriously Peter Piper Pizza style. The claim to be able to increase “focus,” so we have to
design a scenario that will test that. Some will receive the supplement, some won’t.
Maybe they’d do something more academic, but I like the Peter Piper approach.
Step 5 Gather Results
Record the participants scores on a quiz they take over the images they saw and compare between those took the
supplement and those who didn’t.
Some confounding variables could include: innate academic ability, mental state, amount of sleep participants
had the night before, prior familiarity with academic material they were quizzed over, etc.
5. 1200 NAU students are surveyed and 441 said they like the band Radiohead. Create a 95% confidence interval for
this statistic. Interpret your result.
pˆ 
 0.3675
0.3675(1  0.3675)
 .014
So the confidence interval is from 0.37 – 2(0.014) to 0.37 + 2(0.014), or from 0.342 to .398.
This means that I can be 95% confident that the true percent of NAU students who like Radiohead is between
34.2% and 39.8%.