Question 1202151: The high temperatures (in degrees Fahrenheit) of a random sample of 7 small towns are:
96.5
98.7
99.4
97
97.9
98.9
97.8
Assume high temperatures are normally distributed. Based on this data, find the 90% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).
90% C.I. =
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: (97.26, 98.80)
That is the condensed form of 97.26 < mu < 98.80
Work Shown:
mu = μ = Greek letter representing population mean
Specifically mu is the population mean high temperature.
The goal is to estimate mu using a confidence interval.
Given data set = {96.5,98.7,99.4,97,97.9,98.9,97.8}
n = 7 = sample size
xbar = sample mean
xbar = (add up the values)/(number of values)
xbar = (96.5+98.7+99.4+97+97.9+98.9+97.8)/(7)
xbar = 98.029 approximately
Use a calculator or spreadsheet to determine the sample standard deviation is approximately s = 1.045
df = degrees of freedom = n-1 = 7-1 = 6
Because the population standard deviation (sigma) is not known, and because n > 30 isn't the case, we must use the T distribution.
Refer to this T table
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Locate the row labeled df = 6
Locate the column labeled "90% confidence". The confidence labels are at the bottom.
The approximate value t = 1.943 is at this row and column intersection. It is the t critical value.
Specialized stats calculators such as this one
https://www.omnicalculator.com/statistics/critical-value
can find the t critical value.
Make sure to do a two-tailed test.
Also set the significance level to 0.10 (recall that alpha = 1-C where C is the confidence level)
Compute the margin of error
E = t*s/sqrt(n)
E = 1.943*1.045/sqrt(7)
E = 0.767
That result is approximate.
Then,
L = lower boundary
L = xbar - E
L = 98.029 - 0.767
L = 97.262
L = 97.26
and
U = upper boundary
U = xbar + E
U = 98.029 + 0.767
U = 98.796
U = 98.80
The 90% confidence interval in the format L < mu < U is roughly 97.26 < mu < 98.80
That is then condensed to the format (L, U) so we get (97.26, 98.80) as the final answer.
We are 90% confident that the population mean high temperature is somewhere between 97.26 degrees Fahrenheit and 98.80 degrees Fahrenheit.
|
|
|
