Chapter 6 Key, Odd

Review questions   

  1. With evidence against the the null hypothesis.
  2. (A) Hypothesis statements (B) Test statistic (C) P-value (D is optional) Significance conclusion 
  3. The researcher hopes to reject H0. 
  4. The P-value is the probability of the data or data that is more extreme assuming the null hypothesis is  true.
  5. The value for µ0 comes from the research question or an external "norm." 
  6. Because they must consider random differences in either direction, either up or down from the hypothesized value.
  7. SE 
  8. Table:
  Truth

Decision

 H0 True H0 False
Retain H0 Correct retention Type II error
Reject H0  Type I error Correct rejection
  1. beta  = probability of a type II error
  2. 1 - beta  = power 

  3. The criminal justice analogy: 

    Truth
    Decision Innocent Did crime
    Not Guilty  Acquit innocent person Acquit guilty person
    (Analogous to type II error)
    Guilty Convict innocent person
    (Analogous to type I error)    		 Convict guilty person

 

Part A: General Concepts and z Procedures

6A.1. Misconceived hypotheses. What is wrong with each of these statements? 

(A) H0: µ = 100 vs. H1: µ 110. The null and alternative hypotheses must be set up so that either H0 or H1 is true. In this example it is possible for neither to be true. 
(B) H0: = 100 vs. H1: < 100. These hypothesis statements address the sample mean . Hypothesis statements never address sample statistics. They always address population parameters (which in this case should be µ).

6A.3 Satisfaction survey

(A) The Central Limit Theorem. 
(B) The standard deviation of = SEM =  7.5 / 36 = 1.25
(C) ~ N(50, 1.25). Curve is centered on µ = 50 with standard error landmarks at 47.5, 48.75, 50.0, 51.25, and 52.5.
(D) The mean score of 48.4  is not too far from the middle of the distribution. 

6A.5 Satisfaction survey.

(A) zstat = (48.8 – 50) / 1.25 = –0.96. P = Pr(zstat < –0.96) = 0.1685 [from Z table]. This is not significant at α = 0.05 [P > α]
(B) zstat = (46.5 – 50) / 1.25 = –3.50. P = Pr(zstat < –3.50) ≈ 0.0002 (actually a little less than 0.0002). This is significant at α = 0.05 and α = 0.01 [P < α].

6A.7 Lithium

(A) H0: µ = 1.3  vs. H1: µ 1.3; SEM = 0.3 / 25 = 0.06;  zstat = (1.4 - 1.3) / 0.06 = 1.67; P = 2 × Pr(zstat >= 1.67)  = 0.0950. This provides marginally significant evidence against H0.
(B) The results are significant at alpha = 0.10 but not at alpha = 0.05. The jargon to express this level of significance is "marginally significance."

6A.9 NHES. 

(A) SE = 90 / 36 = 15
(B) Sketch the curve ~ N(210, 15). A value of 240 is 2 standard deviations above the hypothesized µ;  Pr( >= 240) = Pr(Z >= 2) = 0.0228. [The sketch of the SDM is not presented because of of the difficulty rendering the graphic on the web. The SDM is centered on 210 and has landmarks 165, 180, 195, 210, 225, 240, 255, starting 3 standard errors below the mean. These correspond to z scores -3, -2, -1, 0, 1, 2, 3. The observed sample mean of 240 is 2 standard errors above the expected mean.] 

6A.11 Fathers had heart attacks. Using a two-sided alternative hypothesis, determine whether the sample mean is significantly different than expected. [Show all hypothesis testing steps.] 

(Step A) H0: µ = 175 mg/dl  vs. H1: µ  175 mg/dl 
(Step B) SEM = 50 / 39 = 8.01; zstat = (195 - 175) / 8.01 = 2.50 
(Step C) Pr(Z 2.50) = 0.0054. Since this is a two-sided test, P = 2  × 0.0054 = 0.0108 
(Step D) The evidence against H0 is significant

6A.13 LDL & fiber. The P-value lets you know that the probability the observed difference occurred by chance is only 1 in 100; this is an unlikely explanation for the results.

Part B: t Procedures

6B.1. P-value from tstat. A test of H0: µ = 0  based on n = 16 calculate tstat = 2.44.

(A) How many degrees of freedom are associated with the test statistic? 15 
(B) Provide the t quantiles from the t table that bracket the tstat. 2.13 and 2.60 
(C) What are the right-tail probabilities for the bracketing t quantiles? 0.025 and 0.01
(D) What is the one-sided P-value for this problem? 0.01 < P < 0.025  
(E) What is the two-sided P-value?  0.02 < P < 0.05 

 

6B.3. Beware a = 0.05. 

(A) No, you would not reject H0 since P > a.  
(B) Yes, you would reject  H0 since P  < a.  
(C) It is not reasonable to come to a different conclusion. The results are nearly identical.

6B.5. Menstrual cycle length. Test whether the mean length of the menstrual cycle in this population is a lunar month. (A lunar month is 29.5 days.) 

 

(Step A) H0: µ = 29.5 days vs. H1: µ  29.5 days
(Step B) n = 9, = 27.78, s = 2.906, sem =  0.9687, tstat = (27.78 - 29.5) / 0.9687 = -1.77; df = 9 - 1 = 8; 
(Step C) The one-sided P value region is between 0.05 and 0.10. The two-sided P value is between 0.10 and  0.20 (P = 0.11 by computer).
(Step D)  The evidence is not significant; we would retain H0 for want of evidence. 

6B.7. Cholesterol levels in Asian immigrants H0: µ = 190 vs. H1: µ 190; : tstat = (181.52 - 190) / (40 / 100) = -2.12; df = 100 - 1 = 99 We will use the t distribution with df = 100 as the best approximation to t99.  The one-sided P-value is between 0.025 and 0.01 and the two-sided P-value is less than 0.05 and more than 0.02.

6B.9 SIDS. H0: µ = 3300 grams vs. H1: µ  3300 grams; x-bar = 2890.5, s = 719.5. SE = 227.62; tstat = -1.80, df = 9, P = 0.106; the mean is not significantly different than 3300 [A non-significant t test is not evidence that the means are equal, especially when the sample is small.]