Question

1a) Assume that a sample is used to estimate a population mean μμ. Find the margin of error M.E. that corresponds to a sample of size 9 with a mean of 68.3 and a standard deviation of 15.8 at a confidence level of 95%.

Report ME accurate to one decimal place because the sample statistics are presented with this accuracy.
M.E. = ?

(Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.)

1b) Assume that a sample is used to estimate a population mean μμ. Find the 80% confidence interval for a sample of size 783 with a mean of 43.4 and a standard deviation of 13.3. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

_?_ < μμ <_?_

(Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.)

Answer 1

Answer 1a)

Margin of Error (ME) = tc*s/sqrt(n)

Here, n = 9 and s = 15.8

Level of Confidence = 95%
α = 100% - (Level of Confidence) = 5%
α/2 = 2.5% = 0.025

Calculate tα/2 by using t-distribution with degrees of freedom (DF) as n - 1 = 9 - 1 = 8 and α/2 = 0.025 as right-tailed area and left-tailed area.

Critical t value tc = tα/2 = 2.306 (Obtained using critical t value calculator)

ME = (2.306*15.8)/sqrt(9)

ME = 12.1449

M.E. = 12.1 (Rounded to one decimal places)

Answer 1b)

80% confidence interval is 42.8 < μ < 44.0