Question

PPH4801/MPHDHM9 Jan/Feb 2018 12 A random sample of 300 diastolic blood pressure measure blood pressure is 99% confidence mmH g lf a 95% confidence interval is also calculated, then interval for the population mean diastolic blood pressure iS 68 to rs The 95% confidence interval will be wider than the 99% The 95% confidence interval will be narrower than the 99% 95% and 99% confidence interval will be the same 4 One cannot make general statement about whether the 95% confidence interval would be narrower, wider or the same as the 99% 13 In a health care utlzation yournal, results are reported from a study peformed on random sample of 100 delvenes at a large teaching hospital The s ample mean birth s reported as 120 grams, and the sample standard deviaton is 25 grams The to report a 95% confidence interval for the population birth weight researchers neglected at is mean birthweight for all delivenes in the hospital) You decade to so, and find the 95% confidence interval for the population mean birth weight to be 1 119 5 grams to 120 5 grams 2 115 grams to 125 grams 3 70 grams to 170 grams 4 1175 grams to 122 5 grams 14 A survey was conducted on a random sample of 1000 Baltmore residents Residents were asked whether they have health insurance 650 individuals surveyed said they do have a health insurance, and 350 said they do not have a health insurance. A 95% Cl for the proportion of Baltımore residents with health insurance is 60% to 75% 32% to 38% 62% to 68% 36% to 48% 3 15 Measurements on babies of mothers who used marjuana dunng pregnancy were compared to measurements on babies of mothers who did not The sample mean head circumference was larger in the group who were not exposed to mar uana and the 95% confidence interval for the difference in mean cireumference between the 2 groups was 0 61 to 1.19cm What can be said about the (2 sided) p value for testing the hypothesis of equal means? 1 The p value is greater than 0 05 2 The p value is equal to 0.05 3 The p value is less than 0 05 4 The p value is 0

Answer 1

Solution : ( 12 )

$\mathrm{2.\quad\:The\:\:95\%\:\:confidence\:interval\:will\:be\:narrower\:than\:\99\%}$

$\mathrm{Explanation\:\::}$

$\mathrm{Higher\:the\:confidence\:interval,\:wider\:will\:be\:the\:interval.\:Therefore,\:\:lower\:the\:confidence\:level\:,\:narrower\:will\:be\:the\:interval.}$

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Solution : ( 13 )

$\mathrm{2.\quad\:\:\:\bold{115\:grams\:\:to\:\:125\:grams}.}$

$\mathrm{Explanation\:\::}$

$\bar{x}\mathrm{\: =120}$

$\mathrm{n=100}$

$\mathrm{\sigma =25}$

$\alpha \mathrm{\: =100-CI}=100-95=5$

$\alpha \mathrm{\: =5\%=0.05}$

$\mathrm{Therefore,\: 0.025\:on\:each\:side\:\left ( \frac{0.05}{2}=0.025 \right )}$

$\mathrm{P(Z)}=1-0.025=0.975$

$\mathrm{From\:standard\:normal\:distribution\:table,\quad\:the\:\:Z-value\:is\:\:1.96}$

$\mathrm{Z_{\frac{\alpha }{2}}=1.96}$

$\mathrm{Now,}$

$\mathrm{The\:95\%\:confidence\:interval\:for\:the\:population\:mean\:birth\:weight\::}$

$\bar{x}- \mathrm{Z_{\frac{\alpha }{2}}}*\frac{\sigma }{\sqrt{n}}\quad\: \mathrm{to}\quad\:\bar{x}+ \mathrm{Z_{\frac{\alpha }{2}}}*\frac{\sigma }{\sqrt{n}}$

$=120-1.96* \frac{25}{\sqrt{100}}\quad\: \mathrm{to}\quad\:120+1.96* \frac{25}{\sqrt{100}}$

$=120-1.96* \frac{5}{2}\quad\: \mathrm{to}\quad\:120+1.96* \frac{5}{2}$

$=120-4.9\quad\: \mathrm{to}\quad\:120+4.9$

$=115.1\quad\: \mathrm{to}\quad\:124.9$

$=115\quad\: \mathrm{to}\quad\:125$

$\mathrm{Hence,\:\:the\:\:95\%\:confidence\:interval\:for\:the\:population\:mean\:birth\:weight\:is\:\:\bold{115\:grams\:\:to\:\:125\:grams}.}$