Basic Applied Statistics 200
Solutions to Midterm 2
-
- P(X>15)=P(Z>(15-11)/2)=P(Z>+2)=P(Z<-2)=.0228
- Half of a normal variable's values fall below the mean: 11
[or take .5 is the probability of being below z=0 and unstandardize
to 11+0(2)=11]
-
- mean of all sample PROPORTIONS is population proportion .17, and sd
is square root of (.17)(1-.17)/75=.043 [quite a few students confused
formulas for counts and proportions; I didn't take off full points for
this mistake]
- In the long run, 95% of the 100 intervals, or 95, should contain p.
- In the long run, 5% of the 100 tests, or 5, should reject a true Ho.
-
- 1784
- for approximate binomial, there must be approximate independence, and
the population should be at least 10 times the sample size: at least
10(1784)=17840
- for approximate normality, check np and n(1-p) both at least 10
- COUNT X has mean np=1784(.06)=107 and sd square root of np(1-p) = 10
- P(X>100)=P(Z>(100-107)/10)=P(Z>-.7)=P(Z<+.7)=.7580
- a probability of .7580 would be characterized as "not unusual" [almost
guaranteed is too strong]
-
- 48/400=.12
- Ho:p=.05 vs. Ha:p>.05
- sd is square root of (.05*.95)/400 and z=(.12-.05)/sd=6.42
- P(Z>6.42)=0, approximately
- Since the p-value is extremely small, we reject Ho and conclude there
is convincing evidence that the alternative is true: population proportion
has increased
- For CI, sd is square root of (.12*.88)/400 and the CI is
.12 plus or minus 2.576*sd = .12 plus or minus .04 = (.08, .16)
[According to Table A.2, 2.576 is the z* multiplier for 99% confidence]
-
- (ii) the manufacturer would be most immediately affected by rejecting
a true null hypothesis, because this would mean shutting down production,
even though the catheters are actually fine
- Use the t* multiplier for 9-1=8 df and 98% confidence, which is 2.90:
CI is 2.01 plus or minus 2.9(.052)/3=2.01 plus or minus .05 = (1.96, 2.06)
- (i) is the only correct interpretation
- (i) larger n results in a narrower interval, because its square root
is in the denominator of our expression for the margin of error
-
- Ho: mu = 23.3 vs. Ha: mu > 23.3
- t=(24.2-23.3)/(5.3/square root of 40)=1.07
- For 30 df, 1.07 < 1.70 so P-value > .05 [Note: you must give a RANGE
for the p-value for this and for part (f); it's important to know HOW the
p-value compares to .05---whether it is smaller or larger.]
- (ii) the p-value is NOT small so do NOT reject Ho
- (ii) yes; 40 is large enough that the standardized statistic will still
follow the t distribution even if the population is non-normal
- p>2(.05)=.10
- (ii) a statistic called s
-
- (ii) no, it was a two-sample study
- smaller n minus 1 = 21-1=20
- (i) we definitely reject the null hypothesis of equal mean sugar
contents, because the p-value is close to zero
- (iii) one quantitative (sugar content) and one categorical (adult or
children's)
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