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 in any calculations and round non-integer results to 4 decimal places.
1. What is the sample mean? 2115
2. What is the margin of error? 85
3. What is the sample standard deviation?
3)
t value at 95% = 2.101
ME = t *SE
85 = 2.101 * SE
SE = 40.4569
SE = s/sqrt(n)
40.4569 = s/sqrt(19)
std.dev = 176.3476
A news report states that the 95% confidence interval for the mean number of daily calories...
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 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 90% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1930, 2170). 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 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...
(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...
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,...
Confidence Intervals 9. Construct a 95 % confidence interval for the population mean, . In a random sample of 32 computers, the mean repair cost was $143 with a sample standard deviation of $35 (Section 6.2) Margin of error, E. <με. Confidence Interval: O Suppose you did some research on repair costs for computers and found that the population standard deviation, a,- $35. Use the normal distribution to construct a 95% confidence interval the population mean, u. Compare the results....
Confidence Interval Given. Assume I created a 95% confidence interval for the mean hours studied for a test based on a random sample of 64 students. The lower bound of this interval was 4 and the upper bound was 14. Assume that when I created this interval I knew the population standard deviation. Using this information, (a) Calculate the width of the interval. (b) Calculate the margin of error for the interval. (c) Calculate the center of the interval. (d)...