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1. Among 734 randomly selected Internet users, it was found that 340 of them use the
Internet for making travel plans (based on data from a Gallup poll). Use a 0.01 significance level to test the claim that among Internet users, less than 50% use the Internet for making travel plans. (12 points)
Reject H0 if T.S. is less than critical value -2.33 (table A) or if p-value is less than 0.01.
p-value is area to the left of -2.0048 = .0228 > .01 = α So Accept Ho. Conclude, the evidence does not support the claim that less than 50% of
internet users use the internet for making travel plans. It appears that it is more likely that roughly 50% use the internet for such a purpose.
2. In clinical experiments involving different groups of independent samples, it is important
that the groups be similar in the important ways that affect the experiment. In an experiment designed to test the effectiveness of paroxetine for treating bipolar depression, subjects were measured using the Hamilton depression scale with the results given below (based on data from “Double-Blind, Pacebo-Controlled Comparison of Imipramine and Paroxetine in the Treatment of Bipolar Depression,” by Nemeroff et al., American Journal of Psychiatry, Vol. 158, No 6). A higher score means a person is depressed, above 17 means the person has some form of depression.
GROUP 1 = Paroxetine treatment group

*n *= 31,

*x *= 18.38,

*s *= 3.91
GROUP 2 = Placebo Group

*n *= 41,

*x *= 21.57,

*s *= 3.87
a. Test the claim that the Paroxetine group has a lower mean score on the Hamilton
scale than the Placebo group. Use a 0.05 significance level. (12 points)
Reject Ho if TS is less than critical value = -1.697 (table B)or if p-value is less than alpha.

Since p-value = .000505 < .05 reject Ho. Conclude the evidence does support the claim that the Paroxetine group has a lower mean score on the Hamilton scale than the Placebo group. (So the drug does appear to work in helping the subjects to be less depressed, or at least score lower on the test that measures their level of depression)
3. The drug Clarinex is used to treat symptoms from allergies. In a clinical trial of this
drug, 3.1% of the 1655 treated subjects experienced fatigue. Among the 1652 subjects given placebos, 1.2% experienced fatigue (based on data from Schering Corporation). Use a 0.05 significance level to test the claim that the incidence of fatigue is greater for those who use Clarinex. Also find 99% confidence interval for the difference in two proportions. What can you conclude from your confidence interval? (12 points)
Group1 – Clarinex Group2 = placebo n1 = 1655 n2 = 1652

*x *=

*p n *= .031(1655) = 51.3 ≈ 51

*x *=

*p n *= .012(1652) = 19.8 ≈ 20
Reject Ho if TS is greater than critical value of z = 1.645 or p-value is less than .05 (alpha)
Since p-value is .000103 < .05 so reject Ho, the evidence does support the claim that the
proportions of those suffering fatigue is greater for those using Clarinex.

99% confidence interval : (.031−.012) ± 2.575
We are 99% confident that the true difference in the proportion of those who suffer fatigue
using Clarinex and those not using Clarinex is between .6% and 3.2%. Because the interval does not contain zero we can conclude that there is a difference in the two proportions. The interval is completely positive, we may infer that the Clarinex users suffer more fatigue.
4. A study was conducted of babies born to mothers who use cocaine during pregnancy. A
sample of 91 babies had a mean weight of 2700 grams with a standard deviation of 645 grams (based on data from “Cognitive Outcomes of Preschool Children with Prenatal Cocaine Exposure,” by Singer et al.,

*Journal of the American Medical Association* Vol. 291m No 20). Find a 95% confidence interval for the mean weight of babies born to mothers use used cocaine during pregnancy. Can you conclude that the mean weight of the babies is less than 2900 grams which is the mean weight for babies born to mothers who do not use cocaine? (12 points)
(the df = 90, but there is no 90 on table so I choose df = 100, you could also have chosen df =80)
We are 95% confident that the true mean weight of the babies born to mothers who used cocaine during pregnancy is between 2565.9 grams and 2843.1 grams Since the interval is entirely below 2900, the mean weight of normal babies we may conclude that the babies born to mothers who used cocaine is lower.

5. Listed below are the heights (in inches) of mother’s and heights (in inches) of their
daughters (based on data from the National Health Examinations Survey”.

**NOTE**: This is the same data set as the previous question.

Mother’s

63.1 67.1 64.7 60.8 65.9 67.1 59.1 60.9 61.4
height = Before Daughter’s 58.6 65.7 65.3 61.0 65.4 67.4 60.9 63.1 62.3 Height = After L1-L2 4.5 1.4 -.6 -.2 .5 -.3 -1.8 -2.2 -.9
Test the claim that there is no difference in the height of the mother and daughters. Use 0.05
significance level. You may use either the p-value or traditional method.
Find a 90% confidence interval for the mean difference in the mothers and daughters heights. Write a conclusion sentence. Does the interval contain zero? What is the significance of the interval containing the value of zero? (12 points)
L1-L2 sto L3 to get the differences into L3
Reject Ho if test statistic is less than -2.306 or greater than 2.306. or if p-value is less than .05.
p-value = .948 > .05 so accept Ho. The evidence supports that claim (Ho) that there is no difference in the mother’s and daughter’s heights. All mother daughters pairs have about the same height.
= (-1.1898, 1.2778), we are 90% confident that the true mean
difference in the mother’s and daughter’s heights is between -1.2 and 1.3 inches. Because zero is in the interval we may conclude that there is no difference in the heights of mothers and their daughters.

6. A group of desperate future law school students all paid $5000 for a prep-course for the
LSATs (the exam all students have to take to get into law school). Since all of the students really want to go to top tier law schools they felt it was worth the money to try to do better on the exam. Listed below are their before and after scores.
Before 140 142 148 154 159 160 160 165 10 172 After 160 154 156 152 150 140 150 139 140 130 B
a. At the 0.05 significance level that the Before score is different from the after
b. Find a 95% confidence interval for the difference in the Before and After scores.
Write a conclusion sentence and indicate the significance of zero being (or not) in the interval. (6 points)
L1-L2 sto L3 to get the differences into L3
Reject Ho if test statistic is less than -2.262 or greater than 2.262. or if p-value is less than .05.
Since the p-value is .149 > .05 we accept Ho, and conclude that there is no difference in the before and after scores. The course does not appear to make a difference in the persons score.
, (-4.31, 24.11). We are 95% confident that the true mean difference in
the before and after scores is between 4.32 and 24.11. Because zero is in the interval we can conclude that there is no difference between the before and after scores.

Source: http://homepage.smc.edu/mcgraw_colleen/math_52/Chapter%2010%20examples%20part%20b%20answers.pdf

DEN RELATIVA RISKEN FÖR HJÄRT/KÄRLHÄNDELSER MINSKADE MED NÄRA HÄLFTEN HOS KVINNOR SOM BEHANDLADES MED ROSUVASTATIN JUPITER-studien, som gjordes med rosuvastatin 20 mg, visar som första statinstudie en minskning av risken för hjärt/kärlhändelser hos kvinnor utan tidigare känd hjärt/kärlsjukdom. I den nya analysen av 6.801 kvinnor från JUPITER-studien minskade antalet hjärt/

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