Tratornos alimenticios

BUSIU 2341-02 Statistics Spring 2009 Final Answer Key

1. a. Normal Rank  ‐1.59  ‐1.06  ‐0.73  ‐0.46  ‐0.22  0.00  0.22  0.46  0.73  1.06  1.59  Ordered X  3  4  5  6  7  8  9  10  12  15  18 

             

b. Comments: The curved pattern of the plot indicates that the distribution of the data is non-normal. The increasing slope of the pattern from left to right indicates that thedistribution is skewed to the right. 2. a. The marginal probability = (45 + 10)/(45 + 10 + 25 + 5) = .647 b. The conditional probability = 5/(10 + 5) = .333 3. a. Because σ is known, the relevant teststatistic is z and its formula is
z x  0




n

b. The sampling distribution of the test statistic z is standard normal, that is, z is distributed normally with mean zero and variance 1.In our notation: z ~ N(0,1). c. z 
x  0




n



110  115 20 100



5   2 .5 2

d. Because it is a two-tail test, the p-value = P(|z| ≥ 2.5) = 2*NORMSDIST(-2.5) = .0124 e.Decision Rule: Reject H0 if p-value ≤ α = .05. Do not reject H0 if p-value > α = .05.

f. Decision: Reject H0 since .0124 = p-value < α =.05. 4. a. The relevant test statistic is F and its formula is2 s BA F 2 . sCB

b. The sampling distribution of the test statistic is F distribution with numerator degrees of freedom nBA - 1 = 26 -1 = 25 and denominator degrees of freedom nCB - 1 = 16 -1 =15.
2 s BA 48  2 .4 . c. F  2  sCB 20

d. Because it is a two-tail test, the p-value = 2*MIN(FDIST(2.4,25,15), 1- FDIST(2.4,25,15)) = .08 e. Decision Rule: Reject H0 if p-value ≤ α = .01. Do notreject H0 if p-value > α = .01. f. Decision: Do not reject H0 since .08= p-value > α =.01. 5. a. UCL =  3

n 45 25

= 280  3

= 280 + 27 = 307 b. CL =μ = 280

c. LCL =  3


n 45 25= 280  3

= 280 - 27 = 253 6. a. Cp =
USL  LSL 6 (   3 )  (   3 ) = 6

=1 b. Cpk =
|   NSL | 3 3 = 3

=1 7. a. The estimated linear trend model is given by
Sales =...