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Based on a random sample of 1040 adults, the mean amount of sleep per night is...
Based on a random sample of 1160 adults, the mean amount of sleep per night is...
Based on a random sample of 1160 adults, the mean amount of sleep per night is 8.49 hours. Assuming the population standard deviation for amount of sleep per night is 3.6 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (_____,_______). (Round to two decimal places as needed.)
Based on a random sample of 1180 adults, the mean amount of sleep per night is...
Based on a random sample of 1180 adults, the mean amount of sleep per night is 7.85 hours. Assuming the population standard deviation for amount of sleep per night is 1.4 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (DD Round to two decimal places as needed.) Interpret the confidence interval O A. O B. ° C. 0 D. We are 95% confident that the interval actually...
Based on a random sample of 1140 adults, the mean amount of sleep per night is...
Based on a random sample of 1140 adults, the mean amount of sleep per night is 8.56 hours. Assuming the population standard deviation for amount of sleep per night is 1.3 hours, construct and interpret a 90% confidence interval for the mean amount of sleep per night. A 9090% confidence interval is (nothing,nothing). (Round to two decimal places as needed.) Interpret the confidence interval.
Based on a random sample of 1080 adults, the mean amount of sleep per night is...
Based on a random sample of 1080 adults, the mean amount of sleep per night is 8.41 hours. Assuming the population standard deviation for amount of sleep per night is 3.8hours, construct and interpret a 90% confidence interval for the mean amount of sleep per night. A 90% confidence interval is (nothing,nothing). (Round to two decimal places as needed.)
0 Based on a random sample of 1140 adults, the mean amount of sleep per night...
0 Based on a random sample of 1140 adults, the mean amount of sleep per night is 8.42 hours. Assuming the population standard deviation for amount of sleep per night is 2.9 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (2.) (Round to two decimal places as needed.) Interpret the confidence interval. O A. We are 95% confident that the interval actually does contain the true value...
A random sample of 85 students finds that they, on average, they get 6.5 hours of...
A random sample of 85 students finds that they, on average, they get 6.5 hours of sleep a night, with a sample standard deviation of 7 hours. Would you use the z-distribution or the t-distribution to construct a confidence interval? How do you know? Construct a 95% confidence interval for the population mean. How do you interpret this interval? Is it likely that students actually get 7 hours of sleep a night? How can you tell?
Was the A random sample of 40 adults with no children under the age of 18...
Was the A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.96 hours, with a standard deviation of 226 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4 23 hours, with a standard deviation of 155 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between...
6.2.19-T Question Help In a random sample of four microwave ovens, the mean repair cost was...
6.2.19-T Question Help In a random sample of four microwave ovens, the mean repair cost was $85.00 and the standard deviation was $13.00. Assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results. The 99% confidence interval for the population mean μ is (DD (Round to two decimal places as needed.) 6.2.21-T Question Help In a random sample...
A. A random sample of 32 different juice drinks has a mean of 98 calories per...
A. A random sample of 32 different juice drinks has a mean of 98 calories per serving and a standard deviation of 31.5 calories. Construct a 99% confidence interval of the population mean number of calories per serving, and interpret the 99% confidence interval in 1 sentence: B. A random sample of 50 standard hotel rooms in Philadelphia, PA, has a mean nightly cost of $189.99 and a standard deviation of $35.25. Construct a 95% confidence interval of the mean...
1.3.21 Question Help A random sample of 40 adults with no children under the age of...
1.3.21 Question Help 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 mear da y leisure time o 4.02 hours, with a standard deviation o 1.84 hours. Construct and interpret a 95% oonfidence interval or the mean dif erence lnle sure...
Based on a random sample of 1180 adults, the mean amount of sleep per night is 7.85 hours. Assuming the population standard deviation for amount of sleep per night is 1.4 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (DD Round to two decimal places as needed.) Interpret the confidence interval O A. O B. ° C. 0 D. We are 95% confident that the interval actually...
0 Based on a random sample of 1140 adults, the mean amount of sleep per night is 8.42 hours. Assuming the population standard deviation for amount of sleep per night is 2.9 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (2.) (Round to two decimal places as needed.) Interpret the confidence interval. O A. We are 95% confident that the interval actually does contain the true value...
Was the A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.96 hours, with a standard deviation of 226 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4 23 hours, with a standard deviation of 155 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between...
6.2.19-T Question Help In a random sample of four microwave ovens, the mean repair cost was $85.00 and the standard deviation was $13.00. Assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results. The 99% confidence interval for the population mean μ is (DD (Round to two decimal places as needed.) 6.2.21-T Question Help In a random sample...
1.3.21 Question Help 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 mear da y leisure time o 4.02 hours, with a standard deviation o 1.84 hours. Construct and interpret a 95% oonfidence interval or the mean dif erence lnle sure...
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