# S02 Basic Applied Statistics 200 Solutions to Practice Midterm 2

1.
1. "at least 17" means greater than or equal to 17: .14+.08+.02=.24
2. "no more than 13" means less than or equal to 13: .02 +.10=.12
2.
1. (5 pts.) (iii) skewed right: most households will have between 1 and 4 people, but there will be relatively few larger households, resulting in a long right tail
2. (5 pts.) (ii) sample mean for a large sample is approximately normal, by the Central Limit Theorem
3. (10 pts.) P(Xbar < 2.5) = P(Z < (2.5-2.64)/(1.3/10) = P(Z < -1.08) =.1401
3.
1. 31/94 = .33
2. 11/24 = .46
3. no; males preference of .46 was higher than the general preference of .33, also of course higher than the female preference (20/70=.29)
4.
1. (i) binomial, so sampling without replacement doesn't affect independence too much
2. (ii) normal, so sample size is large enough for the Central Limit Theorem to take effect
5. .9(50) = 45 [but I also accepted simply .9 as an answer]
6. To do the normal approximation to binomial, you need to find mu = np =400(.1) = 40 and sigma = square root of np(1-p) = square root of 400(.1)(.9) =6. Then standardize and find the probability: P(X > 50) = P(Z > (50-40)/6) =P(Z > 1.67) = P(Z < -1.67) = .0475
7.
1. (5 pts.) 22 is a statistic, 2 is a parameter
2. (15 pts.) 22 plus or minus 1.645(2)/square root of 3 = 22 plus or minus 1.9 = (20.1, 23.9)
3. (5 pts.) (i) is the only correct interpretation; (ii) is incorrect because chance doesn't apply to the fixed population mean; (iii) is incorrect because our interval is for the mean, not for the values themselves; (iv) is incorrect because the interval estimates population mean, not sample mean
4. (5 pts.) (1.645)(2)/1.5, quantity squared is 4.8, round up to 5
5. (5 pts.) (iii) increasing n decreases m
8.
1. (5 pts.) yes because 21 is in the interval (20.1, 23.9)
2. (10 pts.) Because the alternative has a greater than sign, the P-value is P(Z>.87) = P(Z<-.87) = .1922 [I took off 5 points for finding P(Z<.87), which is .8078]
3. Using the bottom row of Table C (since sigma is given this is a z distribution, not t), .87 is between .841 and 1.036, so the P-value is between .20 and .15.
4. No; the P-value is large, so we have no compelling evidence against the null hypothesis.
5. (ii) captured sharks aren't a random sample, necessarily. (i) is not worrisome because we've learned that physical characteristics like heights and weights (or lengths) are typically normal
9.
1. (i) strong evidence, but the difference may be small
2. (ii) small alpha is called for here, because we'd need very convincing evidence to shatter our faith in a long-accepted null hypothesis
10.
1. (5 pts.)Null hypothesis: mu = 2.4; Alternative hypothesis: mu > 2.4
2. (10 pts.) t = (4.5-2.4)/(2.1 over the square root of 8) = 2.83
3. (5 pts.) Using 7 df, t is between 2.517 and 2.998, so the P-value is between .02 and .01
4. (5 pts.) (i) the P-value is small
11.
1. (ii) B is matched pairs
2. (ii) B controls better for differences among rats
12.
1. (10 pts.) no; the P-value is .49, which is large
2. (5 pts.) .49/2 = .245