Stats 245.3 – Review 1 (2013) - Solutions

  1. The Blood Levels (mg/dl) of 50 Subjects selected at random are displayed in the following table

TABLE:

4.9 / 23.3 / 3.9 / 2.5 / 7.6
5.5 / 3.9 / 1.0 / 5.0 / 4.2
7.6 / 0.7 / 1.6 / 2.2 / 4.0
2.3 / 14.1 / 1.0 / 6.1 / 5.4
1.2 / 4.3 / 4.8 / 0.7 / 4.8
0.7 / 3.9 / 1.5 / 8.0 / 6.5
4.1 / 6.9 / 2.9 / 2.1 / 2.8
1.5 / 2.0 / 1.1 / 10.6 / 2.0
6.7 / 3.2 / 1.6 / 0.7 / 9.0
2.1 / 2.7 / 3.5 / 8.2 / 4.4

Calculate the mean, median, and quartiles, and construct a box-and-whisker plot.

Solution

Median = observation in 25.5 position = average of 25th and 26th obs. = 3.9

Q1 = observation in 13th position = 2.0, Q3 = 13th observation from the top = 5.5

To construct the box-whisker plot we need to compute the fences to check for outliers

Inner Fences: IQR = Q3 – Q1 = 5.5 – 2.0 = 3.5

f1 = Q1 – 1.5 IQR = -3.25 and f2 = Q3 + 1.5 IQR = 10.75

Outer Fences:

F1 = Q1 – 3 IQR = -5 and F2 = Q3 + 3 IQR = 16

Mild outlier - 14.1, Extreme outlier – 23.3

Box-whisker plot

  1. In a sample of 125 experimental subjects, the mean score on a post experimental measure of aggression was 55 with a standard deviation of 5.
  2. Construct a 95% confidence interval for the population mean.

Solution:

95% confidence limits or

  1. Suppose the sample size is 169 and the mean score is 55 with a standard deviation of 5. Construct a 99% confidence interval for the population mean.

Solution:

If n = 169 , then 95% confidence limits or

  1. Suppose you want to construct a 95% confidence interval for mean aggression scores as above, and you can assume that the standard deviation of the estimate is 5. How many experimental subjects do you need for the Error bound of the interval to be no larger than 0.4?

Solution:

95% Error bound is: if

  1. What would the number of experimental subjects have to be if you want to construct a 99% confidence interval with an error bound, B, no greater then 0.4?

Solution:

99% Error bound is: if

  1. Under the same criteria we decide that n should be large enough so that a 95% confidence interval would have this error bound, B of 0.4
  1. The mean weight of 100 men in a particular heart study is 61 kg with a standard deviation of 7.9 kg. Construct a 95% confidence interval for the mean.

Solution:

95% confidence limits or

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  1. The standard hemoglobin reading for normal males of adult age is 15 g/100 ml. The standard deviation is about 2.5 g/100 ml. For a group of 36 male con- construction workers, the sample mean was 16 g/100 ml.
  2. Construct a 95% confidence interval for the male construction workers. What is your interpretation of this interval relative to the normal adult male population?

Solution:

95% confidence limits or . Because this interval is strictly above 15, this indicates the average hemoglobin level for the male construction workers is higher than normal males.

  1. What would the confidence interval have been if the above results were obtained based on 49 construction workers?

Solution:

95% confidence limits or .

  1. Repeat b for 64 construction workers.

Solution:

95% confidence limits or .

  1. Do fixed-level confidence intervals shrink or widen as the sample size increases (all other factors remaining the same)? Explain your answer.

Solution:

The confidence intervals shrink as the sample size increases , ,

  1. What is the half-width of the confidence interval that you would obtain for 64 workers?

Solution:

  1. The mean diastolic blood pressure for 225 randomly selected individuals is 75 mmHg with a standard deviation of 12.0 mmHg.
  2. Construct a 95% confidence interval for the mean.

Solution:

95% confidence limits or .

  1. Change exercise a. to assume there are 400 randomly selected individuals with a mean of 75 and standard deviation of 12. Construct a 99% confidence interval for the mean.

Solution:

95% confidence limits or .

99% confidence limits or .

  1. How many individuals must you select to obtain the error bound, B of a 99% confidence interval no larger than 0.5 mmHg?

Solution:

99% Error bound is: if

  1. Peripheral neuropathy is a complication of uncontrolled diabetes. The number of cases of peripheral neuropathy among a control group of 350 diabetic patients was 127. Among a group of 110 patients who were taking an oral agent to prevent hyperglycemia, there were 32 cases of peripheral neuropathy.
  2. Is the proportion of patients with peripheral neuropathy comparable in both groups? Perform the test at a = 0.05.

Solution:

The test statistic is:

Thus

and

We will reject H0: p1 = p2 in favour of HA: : p1 ≠ p2 if |z| > z0.025 = 1.960.

Since this is not true we cannot reject H0 using a = 0.05.

  1. Construct exact 95% confidence intervals for the proportion of patients with peripheral neuropathy in the medication group and the proportion of patients in the control group.

Solution:

(1 – a)100% confidence limits for p1 are:

(1 – a)100% confidence limits for p2 are:

  1. Construct a 95% confidence interval for the difference between the proportions of patients affected by peripheral neuropathy in the control group and in the medication group.

Solution:

(1 – a)100% confidence limits for p1 – p2 are:

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  1. In a community health survey, respondents' smoking status was classified into three categories (smoker, quitter, never smoker). In addition respondents were asked for a self-assessment of their health statsus. The Table below shows the results for the cross-tabulation of smoking status and health status. Determine whether the relationship is statistically significant at the a = 0.05 level.

Table:Cross-Tabulation of Smoking Status and Self-Reported Health Status

Self-reported health status / Smoking Status
Smoker / Quitter / Never / Total
Excellent / 40 / 100 / 229 / 369
Very good/good / 172 / 189 / 485 / 846
Fair/poor / 61 / 63 / 153 / 277
Total / 273 / 352 / 867 / 1.492

Table: Expected Frequencies

Self-reported health status / Smoking Status
Smoker / Quitter / Never / Total
Excellent / 67.52 / 87.06 / 214.43 / 369
Very good/good / 154.80 / 199.59 / 491.61 / 846
Fair/poor / 50.68 / 65.35 / 160.96 / 277
Total / 273 / 352 / 867 / 1.492

Table: Standardized residuals

Self-reported health status / Smoking Status
Smoker / Quitter / Never
Excellent / -3.35 / 1.39 / 1.00
Very good/good / 1.38 / -0.75 / -0.30
Fair/poor / 1.45 / -0.29 / -0.63

.

Since we conclude that there is a significant relationship between smoking status and health status.

  1. The researcher was interested in determining if there was a difference in aggressiveness between the first born and the second born of twins. A sample of n = 12 twins were selected with each pair being measured for aggressiveness. The data are tabulated below:

Twin Set / Twin#l (First Born) / Twin #2 (Second Born) / di / signed rank
1 / 85 / 88 / -3 / -3
2 / 71 / 78 / -7 / -7
3 / 79 / 74 / 5 / 5
4 / 69 / 63 / 6 / 6
5 / 92 / 96 / -4 / -4
6 / 72 / 72 / 0 / 1
7 / 79 / 64 / 15 / 12
8 / 91 / 89 / 2 / 2
9 / 70 / 62 / 8 / 8
10 / 71 / 80 / -9 / -9
11 / 89 / 79 / 10 / 10
12 / 87 / 75 / 12 / 11
  1. Use the Sign test to determine if there is a difference between aggressiveness in the two twins (Use a =0.05)

Solution:

S = the number of positive di’s = 8

The Binomial distribution with n = 12, p = 0.50

From this table we see that if H0 is true P[S ≤ 2] = 0.0193 = P[S ≥ 10]

Thus if the critical region is {0, 1, 2, 10, 11, 12} then a = 2(0.0193) = 0386

Since S = 8 does not lie in the critical region we accept H0: no difference

  1. Repeat using Wilcoxon’s Signed Rank test .

Solution:

W+ = 5 + 6 + 1 + 12 + 2 + 8 + 10 + 11 = 55

W- = 3 + 7 + 4 + 9 = 23

Examining tables for Wilcoxon’s statistic we will reject H0: no difference, if either W+ or W-is less than 14. Since neither of these are true we accept H0.

  1. Repeat using the t-test .

Solution:

We use the test statistic where

Thus

Now t0.025 = 2.201 for 11 d.f. Thus H0 is accepted.

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