Question 1
Sample mean = (Lower limit + Upper limit) / 2 = (1930 + 2170)/ 2 = 2050
Sample mean = 2050
Question 2
Margin of error = (Upper limit – Lower limit ) / 2 = (2170 - 1930) / 2 = 120
E = Margin of error = 120
Question 3
Sample standard deviation S is given as below:
S = E*Sqrt(n)/t
We are given 90% confidence level with n = 15, df = n – 1 = 14
So, critical t value by using t-table is given as below:
t = 1.76131
S = 120*sqrt(15)/ 1.76131
S = 263.8706
(3 points) A news report states that the 90% confidence interval for the mean number of...
2) (3 points) A news report states that the 90% confidence
interval for the mean number of daily calories consumed by
participants in a medical study is (2020, 2160). Assume the
population distribution for daily calories consumed is normally
distributed and that the confidence interval was based on a simple
random sample of 20 observations. Calculate the sample mean, the
margin of error, and the sample standard deviation based on the
stated confidence interval and the given sample size. Use...
(3 points) A news report states that the 90% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1750, 1980). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 18 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the...
(3 points) A news report states that the 90% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1920, 2070). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 21 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the...
(3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1860, 2060). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 16 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the...
(3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1750, 1880). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 20 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the...
A news report states that the 95% confidence interval for the mean number of daily calories consumed by participants in a medical study is (2030, 2200). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 19 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution...
Use graphing calculator to explain answer please. Round results
to the nearest 4 decimal places
(3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1870, 2120). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 15 observations. Calculate the sample mean, the margin of error, and the sample...
If
you could include an explanation that would be appreciated!
Problem Value: 3 point(s). Problem Score: 0 %. Attempts Remaining: 3 attempts. (3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1800, 1980). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 21 observations. Calculate the sample mean,...
X6.2.9-TConstruct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed c = 0.90, x̅ = 12.9, s = 4.0, n = 9 The 90% confidence interval using a t-distribution is 6.2.17-T In a random sample of 26 people, the mean commute time to work was 34.8 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 98% confidence interval for the population mean μ...
what is the margin of error and the confidence interval?
Question Help In a random sample of seven people, the mean driving distance to work was 24.7 miles and the standard deviation was 6.6 miles. Assuming the population is normally distributed and using the I-distribution, a 90% confidence interval for the population mean is (15.5, 33.9) (and the margin of error is 9.2). Through research, it has been found that the population standard deviation of driving distances to work is...