# Report Properties
Owner: msnurselwb
Created: Mar 16, 2018
Share: yes
Views: 3259

Results in this report
None

Data sets in this report

Need help?
To copy selected text, right click to Copy or choose the Copy option under your browser's Edit menu. Text copied in this manner can be pasted directly into most documents with formatting maintained.
To copy selected graphs, right click on the graph to Copy. When pasting into a document, make sure to paste the graph content rather than a link to the graph. For example, to paste in MS Word choose Edit > Paste Special, and select the Device Independent Bitmap option.
You can now also Mail results and reports. The email may contain a simple link to the StatCrunch site or the complete output with data and graphics attached. In addition to being a great way to deliver output to someone else, this is also a great way to save your own hard copy. To try it out, simply click on the Mail link.
IHP 525 Module Five Problem Set

IHP 525 Module Five Problem Set

1.       Newborn weight. A study takes an SRS from a population of full-term 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 11-year-old 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.       P-value and confidence interval. A two-sided test of H0: μ = 0 yields a P-value 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 P-value 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 P-value is less than the alpha level of significance. However, 99% confidence interval for population mean include 0. This is because the P-value is above the alpha level of significance, 0.01.

5.       Critical values for a t-statistic. 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 two-tailed z-test at α = 0.05.

a)      In performing a t-test based on 21 observations, what are the critical values for a one-tailed test when α = 0.05? That is, what values of the tstat will give a one-sided P-value that is less than or equal to 0.05? What are the critical values for a two-tailed test at α = 0.05?

The critical values for a one-tailed test are obtained as shown below.

=(1-∝)=1-0.05=0.95

=t (n-1,0.95)=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)=1-0.05⁄2=(1-0.025)=0.975

""

=t(n-1,0.975)=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 two-sided alternative. Show all hypothesis-testing 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=(27.78-29.5)/(2.91/(√9))=-1.7731

Decision; Do not reject the null hypothesis because the absulte t-statistic value is below the t-critical 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 α-level? 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 cavity-free 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 cavity-free 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 D2 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(n-1)))

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)

Data set 1. water fluoridation data set   [Info]