A random sample of 50 students in Holy Angel University shows that they spend an average of 3 hours per day on social media apps with a standard deviation of 0.5 hours. Assume a normal distribution. Construct a 90% and 97% confidence interval for the average number of hours spent on social media apps per day.
Q: clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Find the 95% confidence interval for the population mean wake time of subjects with treatment:…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Obtain the 90% confidence interval estimate of the mean wake time for a population with the drug…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A:
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: The question is about a confidence interval Given : Popl. mean wake time ( μ ) = 101.0 min Randomly…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: We have to find ci.
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: The sample size is 16, the sample mean is 97.8 and the sample standard deviation is 22.6.
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Obtain the 95% confidence interval for the mean difference in leisure time between adults with no…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A:
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Solution: From the given information,
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: The confidence interval is an inference process that can be used to estimate the difference in…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A:
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Given that n1 = 40 , n2 = 40
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Given, A random sample of 40 adults with no children under the age of 18 years results in a mean…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: From the provided information, Sample size (n) = 13 Sample mean (x̅) = 96.5 Standard deviation (s) =…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Assume that µ is the true mean wake time for a population with the drug treatment.
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: given that,mean(x)=5.67standard deviation , s.d1=2.39number(n1)=40mean(y)=4.14standard deviation,…
Q: drug for treating insomnia in older subjects. Before treatment, 13 subjects had a mean wake time of…
A: Given Xbar=95.3 Standard deviation=23.2
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given information After treatment Sample Standard deviation = 20.1 min Sample mean = 78.1 min Sample…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: There are two independent samples which are adults with no children under the age of 18 and adults…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Refer t-distribution table OR use excel function "=T.INV.2T(0.01,12)" to find the critical value of…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Given n1=40 ; x1=5.03 ; σ1=2.46 n2=40 ; x2=4.48 ; σ2=1.52At 90 ; CI , the t is…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia.Before…
A: Obtain the 95% confidence interval estimate of the mean wake time for a population with drug…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: The 99% confidence interval of the mean wake time for a population with the treatment is obtained…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Confidence interval give a range of values for the unknown parameter of the population. The width of…
Q: A study with 81 kids revealed the amount of time they spend watching YouTube Videos has a mean of 65…
A: given data n = 81x¯ = 65s = 10.290% ci for μ = ?
Q: A physician wanted to estimate the mean length of time that a patient had to wait to see him after…
A: Solution: Let X be the waiting time. From the given information, x-bar=30 minutes, s=9 minutes and…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given mean wake time of 22 subjects before treatment is 103 minutes.After treatment 22 subjects have…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Let μa denotes the true population mean for after treatment. Let xb¯ and xa¯ denote the sample mean…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Obtain the 95% confidence interval for the mean difference in leisure time between adults with no…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A:
Q: Sixteen randomly selected business establishments were found to have a mean water consumption of 800…
A: Given that Sample size n =16 Sample mean =800 Standard deviation =175
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given Mean=97.1 Standard deviations=21.3 Sample size=14
Q: A clinic trial was conducted to test the effectiveness of drug for treating insomnia in older…
A: Sample size (n) = 25Sample mean (x̅) = 78.2Standard deviations (s) = 23.6Significance level (α ) =…
Q: For a simple random sample of 40 Honda Accords (4 cylinder, 2.4 liter, 5-speedautomatic), the mean…
A: Sample size , n = 40 Mean gas mileage, Standard deviation,
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A: Here, mean1 is 5.06 and standard deviation1 is 2.36. Mean2 is 4.06 and standard deviation2 is 1.97.…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given that, Population mean, X bar = 98.2 And standard deviation, sigma= 23.3 Population size, n= 24…
Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…
A:
Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Introduction: It is required to find the 99% confidence interval estimate of the mean wake time for…
Q: clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Assume that the 13 sample values appear to be from a normally distributed population.
Q: clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Assume that the 13 sample values appear to be from a normally distributed population.
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- A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 14 subjects had a mean wake time of 103.0 min. After treatment, the 14 subjects had a mean wake time of 75.4 min and a standard deviation of 24.4 min. Assume that the 14 sample values appear to be from a normally distributed population and construct a 99 % confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 103.0 min before the treatment? Does the drug appear to be effective? Construct the 99 % confidence interval estimate of the mean wake time for a population with the treatment. nothing minless than muless thannothing min (Round to one decimal place as needed.) What does the result suggest about the mean wake time of 103.0 min before the treatment? Does the drug appear to be effective? The confidence interval ▼…A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.17 hours, with a standard deviation of 2.45 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.44 hours, with a standard deviation of 1.71 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1-2) Let u, represent the mean leisure hours of adults with no children under the age of 18 and u, represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for (H, -µ2) is the range from hours to hours. (Round to two decimal places as needed.)A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.88 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.29 hours, with a standard deviation of 1.76 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u, - H2). Let u, represent the mean leisure hours of adults with no children under the age of 18 and µ, represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (µ1 - H2) is the range from hours to hours. (Round to two decimal places as needed.)
- A researcher investigated whether people are spending more time watching Netflix or attending Zoom meetings. A group of 45 people who watch Netflix were spending a mean of 11.3 hours per week with a standard deviation of 2.8 hours. A group of 50 people who are attending Zoom meetings were spending a mean of 13 hours per week with a standard deviation of 4.1 hours. Construct a 95% confidence interval for the difference between the two groups’ means. Based on the interval, is there a significant difference between the number of hours people are spending watching Netflix and attending Zoom meetings? If there is a significant difference, which group has the greater mean? Explain. Calculator function:Four hundred badminton enthusiasts burn an average of 250 calories playing for an hour with a standard deviation of 36 calories. Estimate the mean caloriesburned in playing badminton at 99% confidence level.A random sample of the birth weights of 186 babies has a mean of 3103 grams and a standard deviation of 696 grams. These babies were born to mothers who did not use cocaine during their pregnancies. Construct a 95% confidence interval estimate of the mean birth weight for all such babies
- A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 16 subjects had a mean wake time of 104.0 min. After treatment, the 16 subjects had a mean wake time of 78.7 min and a standard deviation of 21.2 min. Assume that the 16 sample values appear to be from a normally distributed population and construct a 99% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective? Construct the 99% confidence interval estimate of the mean wake time for a population with the treatment. nothing min<μ<nothing min (Round to one decimal place as needed.) What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective? The confidence interval ▼ does not include includes the mean wake…A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 16 subjects had a mean wake time of 105.0 min. After treatment, the 16 subjects had a mean wake time of 95.3 min and a standard deviation of 23.5 min. Assume that the 16 sample values appear to be from a normally distributed population and construct a 99% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0 min before the treatment? Does the drug appear to be effective?A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.59 hours, with a standard deviation of 2.32 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.43 hours, with a standard deviation of 1.83 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (H1 -u2). Let µ, represent the mean leisure hours of adults with no children under the age of 18 and u, represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (H1 - H2) is the range from hours to hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? O A. There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours. O B. There…
- A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.75 hours, with a standard deviation of 2.47 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.03 hours, with a standard deviation of 1.78 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u, - H2). Let H, represent the mean leisure hours of adults with no children under the age of 18 and u, represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for (µ1 - H2) is the range from hours to hours. (Round to two decimal places as needed.)A sample of 6 credit ratings for car loan applicants has a mean of 602 and a variance of 4356. Use the data toconstruct a 90% confidence interval for the population standard deviation of credit scores for all applicantsfor car loans. What is the Best point estimate and Include the written statementA random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.01 hours, with a standard deviation of 2.48 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.42 hours, with a standard deviation of 1.68 hours. Construct and interpret a 95 % confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesis .