In a study to evaluate the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with garlic in a processed tablet form. LDL cholesterol levels were measured before and after the treatment. The changes (before – after) in their LDL levels (in mg/dL) have a mean of 3.2 and a standard deviation of 17.5. The task is to construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. The question asks about the implication of the confidence interval regarding the effectiveness of garlic in reducing LDL cholesterol.Links to statistical tables are provided but not described:- Distribution table- Page 1 of the standard normal distribution table- Page 2 of the standard normal distribution tableBelow this introductory information, there's a section where the user is asked to enter the confidence interval estimate of the population mean (μ) in mg/dL, rounded to two decimal places.Following this, there's a question about the implication of the confidence interval for the treatment's effectiveness:Options:A. The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.B. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.C. The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.D. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.There are no graphs or diagrams in the image. **Understanding Positive z Scores**This educational resource provides a table of positive z scores, which is crucial for statistical analysis, particularly in areas involving standard normal distribution.**Graph Explanation:**At the top left, there's a bell-shaped curve representing a normal distribution. The shaded area under the curve starts at zero and extends to the right, illustrating the cumulative probability from the left up to a certain z score.**Table Explanation:**The main table shows the "Cumulative Area from the LEFT" for various positive z scores, ranging from 0.0 to 3.5. The table is structured as follows:- The row headers (vertical axis) represent the z score's integer and first decimal place (e.g., 0.0, 0.1, ..., 3.5).- The column headers (horizontal axis) represent the second decimal place (e.g., .00, .01, ..., .09).- The cell values represent the cumulative probability from the left of the bell curve up to that specific z score.For example:- A z score of 1.2 with a hundredth digit of .03 corresponds to a cumulative probability of .8907.**Critical Note:**- For a z score of 3.49 or greater, consider using .9999 as the cumulative area.**Common Critical Values Box:**- **Critical Values (one-tail)** - .10: z = 1.28 - .05: z = 1.645 - .025: z = 1.96 - .01: z = 2.33 - .005: z = 2.58- **Critical Values (two-tail)** - .20: z = 1.28 - .10: z = 1.645 - .05: z = 1.96 - .02: z = 2.33 - .01: z = 2.58This table and accompanying explanations serve as a guide for understanding how to interpret and utilize positive z scores in statistical analyses.

given data, n=50x¯=3.2s=17.5CI=0.95α=1-0.95=0.05df=n-1=50-1=49tc:=tα/2,df=t0.05/2,49=2.010 [from excel function , code :=TINV(0.05,49)]E=tc*Sn =2.010*17.550 =4.97CI:(X¯ ±E)=(3.2±4.97)=(-1.77,8.17) answer:-1.77 mg/dl <μ<8.17 mg/dl suggest about CI:the confidence interval limits contain 0, suggesting that the garlic treatement did not affect the LDL cholesterol levels.