A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 63 students using Method 1 produces a testing average of 52.2. A sample of 93 students using Method 2 produces a testing average of 72.3. Assume that the population standard deviation for Method 1 is 15.92, while the population standard deviation for Method 2 is 17.96. Determine the 95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 3 of 3 :   Construct the 95% confidence interval. Round your answers to one decimal place.

Q: We have developed a new drug used to reduce fever and want to test its efficacy. Ten ER patients who…

A: First temperature : mean = 101.97 variance = 1.09 Second temperature : mean = 90.14 variance =…

Q: A gardener is testing a new type of carrot. From a sample of 15 carrots, she calculates a mean…

A: Given n=sample size=15 X=mean=7.2 s=standard deviation=1.23

Q: Step 5 of 6:  Identify the level of significance for the hypothesis test.

A: Given that  The population standard deviation is known to be 1.9

Q: Suppose a different police department says their SRS resulted in a 95% confidence interval of 10.1%…

A:

Q: A random sample of 100 households was selected as part of a study on electricity usage, and the…

A:

Q: In a random sample of six people, the mean driving distance to work was 24.9 miles and the standard…

A: The sample size: n=6 Sample mean: x¯=24.9 miles Standard deviation: s=7.9 Confidence level: C=0.99…

Q: In a random sample of eight ​people, the mean driving distance to work was 19.3 miles and the…

A: The question is about confidence interval Given : Randomly selected no. of people ( n ) = 8 Sample…

Q: Random and independent samples of 90 recent prime time airings from each of two major networks have…

A: The sample size for first network is 90, sample mean is 109.2, population standard deviation is 5.1…

Q: In a transportation study of 27 large cities it was found that the number of motor vehicles per…

A: Given : Sample size (n) = 27 Sample mean (X¯) = one.34 i.e 1.34 Sample SD (S) = 0.46 Confidence…

Q: The ages of registered voters in Smith County are normally distributed with a population standard…

A: Given:

Q: Give the best point estimate for the population and construct a 99% confidence interval for the true…

A: Let μ be the population mean score for students. Given that, x¯= 3.4Sample size (n) = 26Sample…

Q: The concentration of benzene was measured in units of milligrams per liter for a simple random…

A: Given the data as   Sample size Sample Mean Sample SD untreated wastewater 7 7.7 1.3 treated…

Q: The quality-control manager of a chemical company randomly sampled twenty 100-pound bags of…

A: It is given that the sample variance is 6.62 and the sample size is 20.

Q: The correlation between a response and an explanatory variable was calculated as part of a simple…

A: From the given data:   Null and Alternate Hypothesis:   H0: ρ=0HA: ρ≠0

Q: The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT)…

A: Since, the population variances are unknown the Student's t-test will be applied. The variances are…

Q: A researcher compares the effectiveness of two different instructional methods for teaching anatomy.…

A:

Q: sample standard deviation of $74.50. a. Develop a  95% confidence interval estimate of the mean…

A: The sample size, i.e., the sampled number of families, n=64.The sample average spent amount(in $)…

Q: Random and independent samples of 90 recent prime time airings from each of two major networks have…

A:

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

A: From the provided information, Sample size (n) = 48 Sample mean (x̅) = 5.4 Sample standard deviation…

Q: The Weston Rescue Service and the Mid-Valley Ambulance Service are tested for response times. A…

A: Provided information is, the Weston Rescue Service and the Mid-Valley Ambulance Service are tested…

Q: A 2016 study reported in Science magazine stated that college math teachers scored a mean of 3.4…

A:

Q: In a test of physical fitness, a fitness trainer for a group of men ages 65 and older from a local…

A: The population mean is known to be 20. Based on the sample evidence you have to test whether the…

Q: A survey conducted by the American Automobile Association (AAA) showed that a family of four spends…

A: Given that: Population mean, μ=215.60 n = 64 x¯=252.45 s = 74.50

Q: Construct a 90% Confidence Interval for the mean.

A: Let  be the population mean of water per day.Given that,Sample size Sample mean Sample standard…

Q: A paper manufacturer has a production process that operates continuously throughout an entire…

A:

Q: An advertising agency is interested in learning how to fit its commercials to the interests and…

A: Given data: Significance level = 0.05 Sample size: n1= 41 and n2 = 41 Claim: There is a significant…

Q: in order to evaluate two methods of teaching statistics students were randomly assigned to two…

A:

Q: The ICU Admission dataset includes a variable indicating the age of the patient. Find and interpret…

A:

Q: A popular weight loss program claims users will lose weight each month. Linda samples 25 dieters and…

A: We have given that,  Sample mean (x̄) = 10 pounds, standard deviation (s) = 2 pounds and sample size…

Q: The International Air Transport Association surveys business travelers to develop quality ratings…

A: Givensample size(n)=50Mean(x)=3.34standard deviation(s)=2.163confidence interval = 95%

Q: Find a 99% confidence interval for μ.

A: The sample size is n = 7 Sample mean = 46.14 Sample SD s = 1.19 At 99% confidence level and df = (n…

Q: A survey of credit card holders revealed that Americans carried an average credit card balance of…

A: M1=$3900M2=$3300n1=400n2=450σ1=$880σ2=$810

Q: Compare risk factors in men and women. Complete the following table to compare men and women using…

A: Given : Data is collected for Men and women at the first examination for each participant (n = 4434)…

Q: For a random sample of 96 smokers, the mean amount of short-term absenteeism from work was 2.15…

A:

Q: A NAPA Auto Parts supplier wants information about how long car owners plan to keep their cars. A…

A: From the provided information, Sample size (n) = 41 Sample mean (x̅) = 7 Sample standard deviation…

Q: company manufactures a certain over-the-counter drug. The company samples 80 pills and finds that…

A: We have given that  Sample size n = 80 Sample mean = 325.5 Standard deviation s=10.3 90%confidence…

Q: A study of 100 patients with back pain reported that the mean duration of pain was 12 months. It is…

A: In question, Given that a study of 100 patients with back pain reported that the mean duration of…

Q: A survey conducted by the AAA showed that a family of four spends on average $215.60 per day while…

A: It is given that sample mean is 252.45 and the standard deviation is 74.50.

Q: A small business wanted to estimate the mean income from their customers. They took a sample of 256…

A: We have to find  the 90% confidence interval estimate for the population mean income of their…

Q: Random and independent samples of 80 recent prime time airings from each of two major networks have…

A: Given n1=n2=80 S1=4.9 S1=4.6

Q: The average expenditure on Valentine's Day was expected to be $100.89. The average expenditure in a…

A: Obtain the 99% confidence interval for the difference between the two population means. The 99%…

Q: Intellectual development (Perry) scores were determined for 21 students in a first-year,…

A: Given data :  n = 21 Average  students  = 3.27 SD = 0.40 9% of CI

Q: We have developed a new drug used to reduce fever and want to test its efficacy. Ten ER patients who…

A: Given Information: Ten ER patients who are running a fever agree to take the new drug instead of the…

Q: In an old study, the weight of a certain breed of pig had a standard deviation of 3.8 pounds. What…

A: Solution

Q: A researcher wants to compare the effectiveness of two different study techniques for a test. The…

A: To calculate the 95% Confidence Interval for the difference in mean test scores for the two…