### 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…

Answered: In a test of the effectiveness of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Do men take less time than women to get out of bed in the morning? The 48 men observed averaged 7.7 minutes to get out of bed after the alarm rang. Their standard deviation was 1.8. The 55 women observed averaged 8 minutes and their standard deviation was 1.9 minutes. What can be concluded at the αα = 0.01 level of significance? For this study, we should use Select an answer t-test for the difference between two dependent population means z-test for the difference between two population proportions t-test for a population mean t-test for the difference between two independent population means z-test for a population proportion The null and alternative hypotheses would be: H0:H0: Select an answer μ1 p1 Select an answer > ≠ = < Select an answer p2 μ2 (please enter a decimal) H1:H1: Select an answer μ1 p1 Select an answer > < = ≠ Select an answer p2 μ2 (Please enter a decimal) The test statistic ? t z = (please show your answer to 3 decimal places.) The…
In a test of the effectiveness of garlic for lowering cholesterol, 36subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9and a standard deviation of 2.28. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0.What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. a) What are the null and alternative hypotheses? b) Determine the test statistic. c) Determine the P-value. d) (reject/do not reject). There is (sufficient/not sufficient) evidence to conclude that the mean of the population of changes are (=, not, >, <) to 0.
In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 17,1. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table, Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? O mg/dL
In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.6 and a standard deviation of 1.54. Use a 0.10 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Do men take a different amount of time than women to get out of bed in the morning? The 54 men observed averaged 4.3 minutes to get out of bed after the alarm rang. Their standard deviation was 1.4. The 59 women observed averaged 5 minutes and their standard deviation was 2.8 minutes. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use t-test for the difference between two independent population means b. The null and alternative hypotheses would be: Ho: μ1 = + μ2 (please enter a decimal) H₁: μ1 # ♦ μ2 (Please enter a decimal) (please show your answer to 3 decimal places.) c. The test statistic t = d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is α f. Based on this, we should fail to reject + the null hypothesis.
In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.8 and a standard deviation of 24.1. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. H,: u=0 mg/dL H;: µ> 0 mg/dL O B. Ho: µ>0 mg/dL H1: µ< 0 mg/dL OC. H,: µ= 0 mg/dL O D. H,: µ= 0 mg/dL H1: µ<0 mg/dL H,: u#0 mg/dL Determine the test statistic. 0.20 (Round to two decimal places as needed.) Determine the P-value. (Round to three…
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In a test of the effectiveness of garlic for lowering cholesterol, 36subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.3 and a standard deviation of 2.39. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
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In a test of the effectiveness of garlic for lowering cholesterol, 64 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of 23.9. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? Ο Α. Ho: μ=0 mg/ldL H1: µ > 0 mg/dL B. Ho : μ=0 mg/dL H1: µ0 mg/dL H1: µ<0 mg/dL D. Ho: µ=0 mg/dL H1:µ#0 mg/dL
In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9 and a standard deviation of 16.7. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses?
In a test of the effectiveness of garlic for lowering cholesterol, 36 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.7 and a standard deviation of 2.33. Use a 0.01 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

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