back to Math 104 page
back to Walter Stromquist's page
1. Null hypothesis: The (population) average is really 750 hours with SD 300 hours.
Alternative hypothesis: The average is something different from 750.
SE = 300 divided by square root of 36 = 50 hours
Note: "300" describes the variation of the light bulbs themselves. "50" describes the variation of sample averages---the act of averaging 36 independent values reduces the variation by this much.
z = -1.00 (which isn't so unusual)
P-value (two-sided) = 0.32 (either look up -1.00 to get .1587 and double it, OR use the 68-95 rule)
This isn't statistically significant by any reasonable standard. We can't reject the null hypothesis. The lightbulbs might be exactly as claimed. Don't start an investigation.
2a. We would expect 10% failures (call it 0.10)
2b. SE = ( 1 / square-root-of(1000) ) times square root of (.10 times .90) = about .01
2c. z = (.16-.10)/(.01) = about +6
2d. P-value is way too small to look up or compute easily. (FYI: Excel says it's about 0.000000001; that's = NORMSDIST(-6.00))
2e. The result is statistically significant. Reject the null hypothesis. Worry about the factory. There is something wrong with the factory or with this experiment, but whatever it is, it isn't random statistical variation.
3. SE = (1 / sqrt(30)) times 1.0 hours = about 0.18
(Use "1.0 hours" there because it is the SD of the sample. We would rather use the SD of the population---effectively, the "box SD"---but we don't know it, so the sample SD will have to do.)
SO: Confidence interval runs from 4.0 minus (2.6 times 0.18) to 4.0 plus (2.6 times 0.18).
(Use 2.6 because that's how many SE's you need on each side to get to 99% confidence.)
The 99% confidence interval works out to: 3.53 to 4.47.
4a. SE = (1/sqrt(400)) times sqrt(.10 times .90) = 0.015 (one and one half percentage points)
4b. For a 95% interval, allow 2 SE's on either side of the reported average:
From 0.10 + 2(0.015) to 0.10 - 2(0.015)
That is, 7 percent to 13 percent.
4c. For reporting margin of error, pollsters do what we just did in 4b but they use 1/2 in place of the sqrt(frac-yes-times-frac-no) term. In this case, they would get
margin of error = (1/sqrt(400)) times 0.5 times 2, so
(that last 2 is because they still need 2 SE's on either side)
margin of error = 5.0 percentage points.
As you see, that's bigger than they really need for Lee's percentage, but that's what pollsters mean by "margin of error."
Undercoverage bias --- no chance of selecting people without land-line phones
Undercoverage bias (or you could call it selection bias or non-response bias) - miss people who don't answer or won't talk to the caller
Selection bias --- Do we know how phone numbers are assigned? Maybe numbers ending in 0 aren't typical of all people. Or, maybe there's something special about the 1000-5000 range.
Response bias --- Maybe a few people aren't answering correctly. We don't know how often this occurs, but it doesn't take many wrong answers to interfere with a percentage as small as 10 percent for Lee
It looks like sampling error is the least of our problems. It wouldn't help much to increase the sample size. Fix the method first.
back to Math 104 page
back to Walter Stromquist's page