Basic Applied Statistics 200
Solutions to Midterm 2

1.  
   1. P(X>15)=P(Z>(15-11)/2)=P(Z>+2)=P(Z<-2)=.0228
   2. 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]
2.  
   1. 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]
   2. In the long run, 95% of the 100 intervals, or 95, should contain p.
   3. In the long run, 5% of the 100 tests, or 5, should reject a true Ho.
3.  
   1. 1784
   2. 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
   3. for approximate normality, check np and n(1-p) both at least 10
   4. COUNT X has mean np=1784(.06)=107 and sd square root of np(1-p) = 10
   5. P(X>100)=P(Z>(100-107)/10)=P(Z>-.7)=P(Z<+.7)=.7580
   6. a probability of .7580 would be characterized as "not unusual" [almost guaranteed is too strong]
4.  
   1. 48/400=.12
   2. Ho:p=.05 vs. Ha:p>.05
   3. sd is square root of (.05*.95)/400 and z=(.12-.05)/sd=6.42
   4. P(Z>6.42)=0, approximately
   5. 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
   6. 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]
5.  
   1. (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
   2. 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)
   3. (i) is the only correct interpretation
   4. (i) larger n results in a narrower interval, because its square root is in the denominator of our expression for the margin of error
6.  
   1. Ho: mu = 23.3 vs. Ha: mu > 23.3
   2. t=(24.2-23.3)/(5.3/square root of 40)=1.07
   3. 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.]
   4. (ii) the p-value is NOT small so do NOT reject Ho
   5. (ii) yes; 40 is large enough that the standardized statistic will still follow the t distribution even if the population is non-normal
   6. p>2(.05)=.10
   7. (ii) a statistic called s
7.  
   1. (ii) no, it was a two-sample study
   2. smaller n minus 1 = 21-1=20
   3. (i) we definitely reject the null hypothesis of equal mean sugar contents, because the p-value is close to zero
   4. (iii) one quantitative (sugar content) and one categorical (adult or children's)

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