IHP 525 Module Five Problem Set
1. Newborn weight. A study takes an SRS from a population of fullterm infants. The standard deviation of birth weights in this population is 2 pounds. Calculate 95% confidence intervals for μ for samples in which:
a) ""n = 81 and = x ̅= 6.1 pounds
b)
=6.1±1.96(2/√81)=(6.1 ±0.436)=(5.664,6.536)
c) ""n = 36 and = x ̅= 7.0 pounds
=6.1±1.96(2/√81)=(6.1""
""""
=7.0 ±1.96(2/√36)=(7.0±0.653)=(6.347,7.653)""
d) ""n = 9 and = x ̅= 5.8 pounds
=5.8±1.96(2/√9)=(5.8±1.307)=(4.493,7.107)""
2. SIDS. A sample of 49 sudden infant death syndrome (SIDS) cases had a mean birth weight of 2998 g. Based on other births in the county, we will assume σ = 800 g. Calculate the 95% confidence interval for the mean birth weight of SIDS cases in the county. Interpret your results.
""
n=49 x ̅=2998 ∂=800
=2998±1.96(800/√49)=(2994±224)=(2774,3222)
The results implies that if repeated samples of infants death were taken and 95% confidence interval was calculated for mean birth weight for each sample, the mean birth weight would lie between 2774 and 3222.
3. Hemoglobin. Hemoglobin levels in 11yearold boys vary according to a Normal distribution with σ = 1.2 g/dL. (a) How large a sample is needed to estimate mean μ with 95% confidence so the margin of error is no greater than 0.5 g/dL? (b) How large a sample is needed to estimate μ with margin of error 0.5 g/dL with 99% confidence?
n=((Z_(∝ )×σ)/(M.E)) = 〖((1.96×1.2)/0.5)〗^2=22.1276
The sample size is 22.
n=((Z_(∝ )×σ)/(M.E)) = 〖((2.58×1.2)/0.5)〗^2=38.3409
The sample size is about 38.
4. Pvalue and confidence interval. A twosided test of H0: μ = 0 yields a Pvalue of 0.03. Will the 95% confidence interval for μ include 0 in its midst? Will the 99% confidence interval for μ include 0? Explain your reasoning in each instance.
When testing the hypothesis, the null hypothesis is rejected if the Pvalue is below the alpha of significance. The alpha level of significance in this question is 0.05. Alternatively, the null hypothesis is rejected if the confidence interval does not contain zero. In this question, the 95% confidence interval for population mean does not contain 0 in its midst. This is because the Pvalue is less than the alpha level of significance. However, 99% confidence interval for population mean include 0. This is because the Pvalue is above the alpha level of significance, 0.01.
5. Critical values for a tstatistic. The term critical value is often used to refer to the value of a test statistic that determines statistical significance at some fixed α level for a test. For example, ±1.96 are the critical values for a twotailed ztest at α = 0.05.
a) In performing a ttest based on 21 observations, what are the critical values for a onetailed test when α = 0.05? That is, what values of the tstat will give a onesided Pvalue that is less than or equal to 0.05? What are the critical values for a twotailed test at α = 0.05?
The critical values for a onetailed test are obtained as shown below.
=(1∝)=10.05=0.95
=t (n1,0.95)=t (20,0.95)=1.725
t_stat ≥1.725
The critical values for a two tailed test are obtained as shown below.
=(1∝⁄2)=10.05⁄2=(10.025)=0.975
""
=t(n1,0.975)=t(20,0.975)=2.086
""
""
t_stat ≥2.086
6. Menstrual cycle length. Menstrual cycle lengths (days) in an SRS of nine women are as follows: {31, 28, 26, 24, 29, 33, 25, 26, 28}. Use this data to test whether mean menstrual cycle length differs significantly from a lunar month. (A lunar month is 29.5 days.) Assume that population values vary according to a Normal distribution. Use a twosided alternative. Show all hypothesistesting steps.
The hypothesis in this case are shown below.
μ=population mean
H_0: μ=29.5
H_1:μ≠29.5
Obtain the critical value
t (8,0.975)=2.306
Obtain the test statistic
t=(x ̅u)/(s/(√n))
Whereby
x ̅=27.78 s=2.91
Therefore,
t=(27.7829.5)/(2.91/(√9))=1.7731
Decision; Do not reject the null hypothesis because the absulte tstatistic value is below the tcritical value (1.7731<2.306).
Conclusion: The mean menstrual cycle length do not differ significantly from a lunar month
7. Menstrual cycle length. Exercise 6 calculated the mean length of menstrual cycles in an SRS of n = 9 women. The data revealed days with standard deviation s = 2.906 days.
a) Calculate a 95% confidence interval for the mean menstrual cycle length.
x ̅=27.78 s=2.906 n=9
t (8,0.975)=2.306
=27.78±2.306(2.906/√9)=(27.78±2.23)=(25.55,30.01)
b) Based on the confidence interval you just calculated, is the mean menstrual cycle length significantly different from 28.5 days at α = 0.05 (two sided)? Is it significantly different from μ = 30 days at the same αlevel? Explain your reasoning. (Section 10.4 in your text considered the relationship between confidence intervals and significance tests. The same rules apply here.)
The confidence interval above contains both values: 28.5 and 30. Therefore, it is possible to conclude that the mean menstrual cycle length is statistically significant different for boh the population means.
8. Water fluoridation. A study looked at the number of cavityfree children per 100 in 16 North American cities BEFORE and AFTER public water fluoridation projects. The table below lists the data. You will need to manually type the data into StatCrunch to use that tool to calculate the requested information.
a) Calculate delta values for each city. Then construct a stemplot of these differences. Interpret your plot.
b) What percentage of cities showed an improvement in their cavityfree rate?
c) Estimate the mean change with 95% confidence.
<data set 1>
The delta values are obtained as shown in the table below.
After 
Before 
D 
D^{2} 
49.2 
18.2 
31 
961 
30.0 
21.9 
8.1 
65.61 
16.0 
5.2 
10.8 
116.64 
47.8 
20.4 
27.4 
750.76 
3.4 
2.8 
0.6 
0.36 
16.8 
21.0 
4.2 
17.64 
10.7 
11.3 
0.6 
0.36 
5.7 
6.1 
0.4 
0.16 
23.0 
25.0 
2 
4 
17.0 
13.0 
4 
16 
79.0 
76.0 
3 
9 
66.0 
59.0 
7 
49 
46.8 
25.6 
21.2 
449.44 
84.9 
50.4 
34.5 
1190.25 
65.2 
41.2 
24 
576 
52 
21.0 
31 
961 
Total 

195.4 
5167.22 
The stem plot for the differences In column 3 are shown below.
STEM 
LEAF 
0 
4 2 0 0 
0 
0 3 4 7 8 
1 
0 
2 
1 4 7 
3 
1 1 4 
Based on the stem plot above, it is clear that the highest change was 34.
The percentage of cities which showed an improvement in their cavity free rate is shown below.
Total number of cities is 16.
Number of cities which D is positive is 12.
a) Estimate the mean change with 95% confidence.
The formula used in this case is shown below.
Average deviation = (∑▒D)/n=195.4/16=12.2125
""
=C.I=d ̅±t_∝ s_d/(√n)
but,
""
""
s_d=√((n∑▒〖d^2(∑▒〖d)^2〗〗)/(n(n1)))
Therefore,
= √((16*5167.22〖195.4〗^2)/(16*15))=3.6159
Confidence interval for the mean difference is obtained as shown below using the information calculated above.
""C.I=12.2125±2.131(13.6159/√16)=12.2125±7.2539=(4.9586,19.4664)
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