5 Key Odd

Review Questions

1. Statistical inference is generalizing from the sample to the population with calculated degree of certainty. 
2. estimation and hypothesis (significance) testing  
3. Two forms of estimation: point estimation and interval estimation. 
4. The goal of estimation is to use sampled data to predict the approximate value of the parameter (via a confidence interval). The goal of testing is to make a judgment about a claim concerning a parameter.
5. Statistics are calculated from the sample and are random variables. Parameters are  population characteristics and are constants.
6. This means that the in the long run, repeated samples of will average µ. 
7. central limit theorem . . . population mean . . . standard error  . . . square root 
8. standard errors
9. True [edited 3/12/07 to make question easier]  
10. Matching: (a) = parameter; (b) = statistical inference; (c) = sampling distribution of the mean; (d) = alpha; (e) = standard error of the mean; (f) = confidence interval
11. 5%
12. alpha × 100%
13. (b) population 
14. margin of error

Exercises Part A 

5A.1 Parameter or estimate? Part 1. Say whether each of the boldface numbers is a parameter or estimate. (A) 18,800 is a parameter. All the other numbers listed in this problem are statistics. (B) Parameter (note: we are assuming that inferences are to be restricted to that year and population.) 

5A.3 Very tiny population. 

(A) Possible samples: {1, 3}, {1, 5}, {1, 7}, {1, 9}, {3, 5}, {3, 7}, {3, 9}, {5, 7}, {5, 9}, {7, 9}. Means from these samples: 2, 3, 4, 5, 4, 5, 6, 6, 7, 8. 
(B) The stemplot of the sampling distribution of means follows. It is more mound shaped (CLT), has the same mean ("unbiasedness"), and is less spread ("square root law") than the population distribution. 

1|
2|0
3|0
4|00
5|00
6|00
7|0
8|0
9|
(x1)

5A.5 Serum cholesterol in undergraduate men

(A) What is the probability of selecting someone at random from this population who has a cholesterol value less than 180? z = (180 - 190) / 40 = -0.25; Pr(Z -0.25) = 0.4013
(B) What is the standard deviation (error) of ? The standard deviation of the sampling distribution of is called the SE and is equal to s / n = 40 / 49 = 5.714.
(C) What is the probability that your (based on a SRS of n = 49) is less than 180? z = (180 - 190) / 5.714 = -1.75; Pr(Z -1.75) = 0.0401

5A.7 Sampling behavior of a mean 

• Draw the sampling distribution of the mean (SDM) based on this sample and identify landmarks on the horizontal axis that are ±2 standard errors around the expected mean. [Note that SEM = 4 / 30 = 0.7303]. ~ N(55.5, 0.7303). The curve is not shown on the web because of technical difficulties. 
• Would you be surprised to find a sample mean less than 54? A mean of 54 has z = (54 - 55.5) / 0.7303 = -2.05. Thus, Pr( 54) = Pr(Z -2.05) = .0202. In words, 2.02% of means will be less than or equal to 54. Since 2.02% is unlikely, we would find this somewhat surprising. 
• Would you be surprised to find a sample mean that exceeds 56?  A mean of 56 has z = (56 - 55.5) / 0.7303 =  0.68. Thus, ( 54) Pr(Z 0.68) =  0.2483. In words, 24.83% of means will be greater than or equal to 56. Since 24.83% is not that rare, this would not be too terribly surprising.

5A.9 Pharmacy survey. Calculate the 95% confidence interval for the mean price of the drug. 95% confidence interval for mu = ± (margin of error) = $33 ± $2.50 = ($30.50 to $35.50) What does it mean to say that we have 95% confidence in this interval? It means that we expected 95% of such intervals to capture mu and 5% to miss it. 

5A.11 Graduate student age. A 95% confidence interval for µ. The SE = 5 / 24 = 1.021. The 95% confidence interval for µ = 25 ± (1.96)(1.021) = 25 ± 2.00 = (23.0, 27.0). 

5A.13 Graduate student age. 99% confidence interval for µ = 25 ± (z_.995)(1.02) = 25 ± (2.58)(1.02) = 25 ± 2.63 = (22.4, 27.6) . This confidence interval is longer because it was constructed to have a greater chance of capturing µ. Thus, z value in the formula (the critical z) is 2.58 instead of 1.95, giving a larger margin of error surrounding the interval.

5A.15 Antigen titer. Given: n = 3, sigma = 0.070; Calculated: = 17.4033, SE = 0.070 / 3 = 0.04041; 95% confidence interval for µ = 17.4033  ± (1.96)(0.0404) = 17.4033 ± 0.0792 = (17.32  to 17.48).

5A.17 SIDS. Calculate a 95% confidence interval for the mean µ birth weight of SIDS cases. 95% confidence interval for µ = 2998 ± (1.96)(800/49)  = 2988 ± 224 = 2764 to 3212. Interpret your results. We have 95% confidence this interval has captured the mean birth weight in the population of SIDs cases. 

5A.19 Hemoglobin study, sample size requirements.  n = (4)(1.209)² / 0.5² = 23.4. Sample size requirements are always rounded up to the next integer to ensure adequate precision. Therefore, 24 people should be sampled.

Exercises Part B 

5B.1. Blood pressure 

(A) Calculate the estimated standard the error of the mean. se = 10.3 / 35 = 1.74 
(B) How many people would you need to study to decrease the standard error of the mean to 1 mm Hg? n = (s / se)² = (10.3 / 1)² = 106.09 → round up to next integer → use n = 107

5B.3. t curve 

(A) Sketch a t curve. Sketches not shown b/c of technical difficulties. 
(B) ... determine the t quantile with 9 degrees of freedom and cumulative probability 0.90... t_9,.90  = 1.38   
(C) Use the t table to determine the value of t with 9 degrees of freedom and cumulative probability 0.90  t_9,.10  = -1.38 Shade the region to the left of this point. Not shown on web. 
(D) What is the combined area of the shaded regions of the curve you just sketched?  0.20 (.1 in the left tail and .1 in the right tail).

5B.5. Approximating the areas beyond a t quantile. The right-tail beyond 2.65 on t₈ is a little bigger than 0.01 and a little smaller than 0.025. The precise probability computed with software = 0.014628.

5B.7.  t score for a confidence interval. Use t_28-1,1-.05/2 = t_{27, .975} = 2.05. 

5B.9.  Red wine and polyphenol levels. Calculate a 95% confidence interval for the mean percent change in polyphenols associated with this amount of red wine consumption. [Calculate: = 5.5 s = 2.517, n = 9] se =  2.517 / 9 = 0.839, df = 9 - 1 = 8. Use t_8,.975 = 2.31. The 95% confidence interval for µ is 5.5 ± (2.31)(0.839) = 5.5 ± 1.9 = (3.6, 7.4) 

5B.11. Boy height. sem = 3.1 / 26 = 0.608, df = 26 - 1 = 25, t_{25, .975} = 2.05 and the 95% CI for µ is 63.8 ± (2.06)(0.608) = 63.8 ± 1.3 = (62.5, 65.1).

5B.13. Respiratory function in furniture workers.  n = 7, = 2.969, s = 0.9876, sem = 0.9876 / 7 = 0.373. Notice that a = .10 for 90% confidence. The  90% CI for µ is  2.969 ± (t_7-1,1-.10/2)(.373) = 2.969 ± (t_6,.95)(.373) = 2.969 ± (1.94)(.373) = 2.969 ± 0.724 = (2.245, 3.693)

5B.15.  Body weight in high school girls. 

(A) Stemplot below shows no major departures from Normality
  9|4
 9|6
10|034
10|55
11|44
×10 (pounds)

(B) x-bar = 103.888, s = 6.918, se = 6.918 / √9 = 2.306, df = 8, 95% CI for µ = 103.888 ± (2.306)(2.306) = 103.888 ± 5.318 = (98.570 to 109.206)  
(C)   margin of error = 5.318

(D) n = (1.96 ∙ 6.918 / 3)² = 20.43 → 21