Respuesta :
Answer:
The 90% confidence interval for the mean net change in LDL cholesterol after the garlic treatment is (-1.7, 7.1).
As there are negative values included in the interval, at 90% confidence, we can not conclude there is no enough evidence about the effectiveness of garlic in reducing LDL cholesterol, as the true mean net change can be 0 or negative.
Step-by-step explanation:
We have to calculate a 90% confidence interval for the mean net change in LDL cholesterol after the garlic treatment.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=2.7.
The sample size is N=47.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{17.8}{\sqrt{47}}=\dfrac{17.8}{6.856}=2.596[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=47-1=46[/tex]
The t-value for a 90% confidence interval and 46 degrees of freedom is t=1.679.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.679 \cdot 2.596=4.4[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 2.7-4.4=-1.7\\\\UL=M+t \cdot s_M = 2.7+4.4=7.1[/tex]
The 90% confidence interval for the mean net change in LDL cholesterol after the garlic treatment is (-1.7, 7.1).
As there are negative values included in the interval, at 90% confidence, we can not conclude there is no enough evidence about the effectiveness of garlic in reducing LDL cholesterol, as the true mean net change can be 0 or negative.
