### Critical t Values#### Graph ExplanationThe image contains a graph depicting the t-distribution curve with a focus on the left tail. The curve illustrates how the t-distribution looks, showing the peak and the tails on either side. Key elements of the graph include:- **Left Tail**: A portion of the curve shaded in green, representing the critical region for negative t values.- **α (Alpha)**: Denotes the significance level, which is the probability of rejecting the null hypothesis when it is true. It's shown near the left tail.- **Critical t Value (Negative)**: Indicated by a line from the horizontal axis to the shaded region, marking the threshold beyond which values are considered significant for a left-tailed test.#### Table: t Distribution - Critical t ValuesThis table contains the critical t values based on varying degrees of freedom and significance levels. It is divided into two main sections: Area in One Tail and Area in Two Tails. The table helps determine the critical value for a specific degree of freedom and significance level.- **Degrees of Freedom (df)**: Listed vertically from 1 to 14.- **Area in One Tail**: Columns include significance levels of 0.005, 0.01, 0.025, 0.05.- **Area in Two Tails**: Columns present significance levels of 0.01, 0.02, 0.05, 0.10.Critical t values are provided for different configurations of significance levels, helping in hypothesis testing:- For example, with 1 degree of freedom and a 0.01 significance level in one tail, the critical t value is 63.657.- With 4 degrees of freedom and a 0.05 significance level in two tails, the critical t value is 2.776.This detailed information is crucial for statistical analysis and understanding the probability of observing extreme values under the null hypothesis. ### Investigating the Effectiveness of Garlic in Lowering CholesterolA study was conducted to test the effectiveness of garlic tablets in lowering LDL cholesterol. The study involved 50 subjects who were administered garlic in a processed tablet form. LDL cholesterol levels were recorded both before and after the administration of the garlic treatment.#### Study Findings:- **Change in LDL Cholesterol Levels:** Subjects showed a mean change of 5.7 mg/dL in their LDL cholesterol levels, with a standard deviation of 16.8.- **Objective:** Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol levels after the garlic treatment. Evaluate the implication of this interval on the effectiveness of garlic in reducing LDL cholesterol.#### Calculating Confidence Interval:What is the confidence interval estimate of the population mean \( \mu \)?\[ \Box \text{ mg/dL} < \mu < \Box \text{ mg/dL} \]*Note: Round to two decimal places as needed.*#### Evaluating the Effectiveness:What does the confidence interval suggest about the effectiveness of the treatment?- **A:** The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.- **B:** The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.- **C:** The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.- **D:** The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels.### Additional Resources:- [View a t distribution table](#)- [View page 1 of the standard normal distribution table](#)- [View page 2 of the standard normal distribution table](#)This exercise aims to help students understand how to construct confidence intervals and interpret their implications in the context of experimental treatments.

Given : Sample Size, n = 50 Sample Mean, x̄ = 5.7 sample standard deviation, s = 16.8 confidence level,c=0.95solution : Significance level, α =1-0.95 = 0.05 Degree of freedom df = n-1 = 49 critical value : critical value at α=0.05 and df=49 is tα/2,df=2.01 .....from student's t distribution table 95% confidence interval =5.7 ±2.01 x 16.8 / √50 =5.7 ±4.775516 (=5.7 -4.775516 ,5.7 +4.775516 ) = (0.92,10.48) 95% confidence interval is 0.92mg/dl<u<10.48 mg/dl b) option b) is correct answer