A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 13 subjects had a mean wake time of 101.0 min. After tréatment, the 13 subjects had a mean wake time of 74.4 min and a standard deviation of 23.5 min. Assume that the 13 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 101.0 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. min < µ < min (Round to one decimal place as needed.) |

Q: A random sample of 4040 adults with no children under the age of 18 years results in a mean daily…

A: To avoid any calculation error , I am using Ti 83 calculator and the command is as follows

Q: A survey of 100 retailers revealed that the mean after-tax profit was $80,000. Assuming that the…

A: Given data,n=100σ=15,000x=80,000z-value at 95% confidence is Zc=1.96

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A: The sample size is 64, sample mean is 96, and standard deviation is 16. The degrees of freedom is,…

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Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…

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Q: A random sample of 4040 adults with no children under the age of 18 years results in a mean daily…

A: In statistical inference, there are two types of estimation,  Point estimation and Interval…

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Q: A study by the department of education of a certain state was trying to determine the mean SAT…

A: Here we need to construct a 95% confidence interval for the population mean. Since the population…

Q: A sample of 352 subscribers to Wired magazine shows the mean time spent using the Internet is 18.4…

A: Given -  Sample size, n = 352 Mean time spent, x=18.4 hoursStandard deviation, σ=6.8 hours

Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…

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Q: A 30 month study is conducted to determine the difference in rates of accidents per month between…

A: Given:x1=12.3s1=3.5n1=12                    and                    x2=7.6s2=3.4n2=9   Since, the…

Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…

A: Introduction: As the sample standard deviations, 2.26 and 1.57, are quite different from one…

Q: A 30 month study is conducted to determine the difference in rates of accidents per month between…

A:

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 consumer watchdog organization estimates the mean weight of 1-ounce "Fun-Size" candy bars to see…

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Q: In a recent study of 42 eighth graders, the mean number of hours per week that they watched…

A: Mean = 19.6 Standard deviation = 5.8 Sample size n = 42

Q: State the conclusion in the problem context.   We fail to reject H0. It is not reasonable to…

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Q: An analyst is planning to estimate the mean of a quantitative variable using a 95% confidence…

A: Given that, Critical value: Using the z-table, the critical value at 95% confidence level is

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 large department store found that it averages 362 customers per hour. Assume that the standard…

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Q: On this recommendation, a researcher wanted to estimate the mean sedentary time (sitting while…

A: Given:- n=25, x¯=8.76,  σ=1.75 To calculate 95% confidence interval of a normally distributed…

Q: A group of 10 surgery patients had a mean weight of 240 pounds with a sample standard deviation of…

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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 report published by the BZW Luxury Association shows that each American household spent on average…

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Q: A random sample of 4040 adults with no children under the age of 18 years results in a mean daily…

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Q: In a manufacturing process, a random sample of 9 manufactured bolts has a mean length of 3 inches…

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Q: A bread shop owner wishes to find the 95% confidence interval of the true mean cost of special…

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Q: In a sample of 15 small candles, the weight is found to be 3.72 ounces with a standard deviation of…

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Q: A study of a local high school tried to determine the mean number of text messages that each student…

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Q: On a particular day a consumer-advice bureau received 125 calls. For a random sample of 40 of these…

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Q: A study by the department of education of a certain state was trying to determine the mean SAT…

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Q: A college alumni Organization sent a survey to all recent graduates to ask their annual income. 20%…

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Q: A study by the department of education of a certain state was trying to determine the mean SAT…

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Q: A study of a local high school tried to determine the mean number of text messages that each student…

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Q: A random sample of 40 adults with no children under the age of 18 years results in a mean daily…

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Q: A study by the department of education of a certain state was trying to determine the mean SAT…

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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.