A study of two kinds of photocopying equipmentshows that 61 failures of the first kind of equipment took on the average 80.7 minutes to repair with a stan-dard deviation of 19.4 minutes, whereas 61 failures of the second kind of equipment took on the average 88.1minutes to repair with a standard deviation of 18.8 minutes. Find a 99% confidence interval for the dif-ference between the true average amounts of time it takes to repair failures of the two kinds of photocopyingequipment.

Q: A randomized controlled trial is run to evaluate the effectiveness of a new drug for asthma in…

A: As per the question we have Drug Placebo n1 = 125 n2 = 125 x̄1 = 12.4 x̄2 = 13.1…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…

A: The random variable LDL cholesterol level follows normal distribution. We have to construct 99%…

Q: A health care company reported that an insured woman spends on average 2.5 days in the hospital for…

A: The degrees of freedom is,…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…

A: Given information- Sample size, n = 48 Sample mean, x-bar = 3.4 Sample standard deviation, s = 18.9…

Q: Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc…

A: Hi! Thank you for the question, as per the honor code, we are allowed to answer three sub-parts at a…

Q: A clinical trial was designed to compare a new drug for lowering cholesterol to a placebo. The 100…

A: Given, A clinical trial was designed to compare a new drug for lowering cholesterol to a placebo.…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…

A: Given that Sample size =44 subjects  Mean difference=3.4  Sample standard deviation (before-after)…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…

A: Here we need to find a 95% confidence interval for the mean net change in LDL cholesterol after the…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…

A:

Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…

A:

Q: and after the treatment. The changes (before-after) in their levels of LDL cholesterol (in mg/dL)…

A: Given n=sample size=49, Sd=17.1,  mean difference x̄d=2.8  Level of significance ɑ=0.05

Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…

A: Obtain the 90% confidence interval estimate of the mean net change in LDL cholesterol after the…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…

A:

Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…

A: Given Information: Sample size n=43 Sample mean x¯d=3.2 Sample standard deviation sd=17.9 Confidence…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…

A:

Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…

A: The random variable net change in LDL cholesterol follows normal distribution. The sample size is…

Q: Consider evidence from a random sample of 15 cars of the same model. The average gas mileage from…

A: given data  assume normal distribution with σ = 5.2sample size (n) = 15sample mean (x¯) = 26.790% ci…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…

A: Let d denotes the changes (before-after) in the levels of LDL cholesterol (in mg/dL) of the…

Q: In a study of 120 post-menopausal women on hormone replacement therapy (HRT), there was a mean…

A: sample size(n)=120Mean ()=131standard deviation (s)=16.5confidence level=95%

Q: Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc…

A: a) Given:- Sample size of data is n=25. Sample mean duration for the zinc lozenge group of cold…

Q: nd a standard deviation of 18.9. Construct a 90% confidence interval estimate of the mean net change…

A:

Q: A randomized controlled trial is run to evaluate the effectiveness of a new drug for asthma in…

A: Let the random variables are as:X1: New drugX2: Placebo

Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…

A: Given n=50 Mean=5.5 Standard deviations=17.9 Alpha=0.01

Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…

A: Given that x¯=5.2s=15.6n=48 Confidence level=90% Significance level=1-0.90=0.10 degrees of freedom,…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…

A: given data sample size (n) = 43sample mean ( x¯ ) = 4.2sample standard deviatio (s) =17.890% ci for…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…

A: Sample mean = x̅ = 4.5 Sample size = n = 43 Sample S.D = s = 16.8 Standard Error = s/√n  =…

Q: Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc…

A: Obtain the 99% confidence interval for the mean duration of symptoms for individuals using the…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 29 randomly subjects were treated…

A: Find the critical values for the chi-square distribution: Refer chi-square table to find the…

Q: (a) Calculate a 95% confidence interval for the mean overall duration of symptoms in a population of…

A:

Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…

A: sample size(n)=49Mean()=3.6standard deviation(s)=17.7confidence level=90%

Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…

A:

Q: Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc…

A: Given that  For Zinc lozenges Sample size=25 Mean=4.5 sample standard deviations=1.4 For Placebo…

Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…

A: Obtain the 99% confidence interval estimate of the mean net change in LDL cholesterols after the…

Q: A sample of 30 randomly selected patients in a waiting room at a doctor’s office reported a mean…

A: Solution: Given information: n= 30 Sample size  x= 38.5 minutes  Sample mean σ= 1.78 minutes…

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.