Question

in a test of effectiveness of garlic for lowering cholesterol, 45 subjects were treated with garlic in process tablet form. chorlesterol levels were measured before and after the treatement. the cahnges in their levels of ldl cholesterol (in mg/dL) have a sample mean 5.1 and a sample standard deviation of 16.7. a.) the best point estimate is _mg/do b.) construct a 99% confidence interval estimate of the mean net change in ldl cholesterol after the garlic treatment. c) interpret this interval

Answer 1

Answer:

Step-by-step explanation:

(a)

The sample statistic which is estimated by the population parameter is the sample mean. Moreover it is the best estimate for the population mean.From the information, it is clear that  

The sample mean is the best point estimate for the population mean. This clearly indicates that the best point estimate for the population mean is 5.1mg/dl.

(b)

From the information, it is clear that  $\bar x = 5.1,\,s = 16.7,\,n = 45$    

The degree of freedom is obtained below:

$df = n - 1\\\\ = 45 - 1\\\\ = 44$  

From “Student’s-t Distribution Table”, the required critical value is 2.704 which is lesser than the critical value.

The 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment is,

${\rm{Confidence}}\,{\rm{interval}} = \bar x \pm {t_{\frac{\alpha }{2},n - 1}}\left( {\frac{S}{{\sqrt n }}} \right)\\\\ = 5.1 \pm {t_{\frac{{0.01}}{2},45 - 1}}\left( {\frac{{16.7}}{{\sqrt {45} }}} \right)\\\\ = 5.1 \pm \left( {2.704} \right)\left( {2.4895} \right)\\\\ = 5.1 \pm 6.7316$

$= \left( {5.1 - 6.7316,5.1 + 6.7316} \right)\\\\ = \left( {-1.6316,11.8316} \right)$  

(c)

From the information, it is clear that the cholesterol levels were measured and identified whether the garlic treatment affects the LDL cholesterol levels or not.

It is clear that, the 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment is $\left( {-1.6316,11.8316} \right)$ . By the rejection rule for the confidence interval it is clear that the confidence interval contains 0 which clearly states that the garlic treatments do affect the LDL cholesterol levels.

Answer 2

Answer:

A) the sample mean is the best point estimate 5.1mg/do

B) the required confidence  interval = [ 0.92 , 9.28]

C) the confidence interval suggest that there is a 99% chance that the true value of the mean net change in LDL cholesterol after garlic treatment is contained in the interval

Step-by-step explanation:

information given from the question

The sample size (n) = 45

The mean of the changes ( before and after ) in the levels of the LDL cholesterol = 5.1

the standard deviation = 16.7

B) the 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment can be constructed using the formula

=   $[ d - t\frac{\alpha }{2}, _{n-1} \frac{sa}{\sqrt{n} },d +t_{\frac{\alpha }{2},n-1 }\frac{sa}{\sqrt{n} }$

d = sample mean = 5.1

sa = standard deviation = 16.7

$t_{\frac{\alpha }{2},n-1 }$ = 1.6787

= $[5.1 - 1.6787 * \frac{16.7}{\sqrt{45} },5.1 + 1.6787 * \frac{16.7}{\sqrt{45} }]$

= [ 5.1 - 1.6787 * 2.4895, 5.1 + 1.6787 * 2.4895 ]

= [ 5.1 - 4.1791, 5.1 + 4.1791  ]

the required confidence  interval = [ 0.92 , 9.28]

C) the confidence interval suggest that there is a 99% chance that the true value of the mean net change in LDL cholesterol after garlic treatment is contained in the interval