SOLUTION: The high temperatures (in degrees Fahrenheit) of a random sample of 9 small towns are:
99.2
98.5
99.7
98.3
97.9
97.8
98.9
96.9
98.1
Assume high temperatures are norma
Algebra ->
Probability-and-statistics
-> SOLUTION: The high temperatures (in degrees Fahrenheit) of a random sample of 9 small towns are:
99.2
98.5
99.7
98.3
97.9
97.8
98.9
96.9
98.1
Assume high temperatures are norma
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Question 1207303: The high temperatures (in degrees Fahrenheit) of a random sample of 9 small towns are:
99.2
98.5
99.7
98.3
97.9
97.8
98.9
96.9
98.1
Assume high temperatures are normally distributed. Based on this data, find the 95% 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).
95% C.I. =
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.
You can put this solution on YOUR website! the mean of the sample = 98.366666667.
the standard deviation of the sample = 0.832165849.
the sample size is equal to 9.
t-score is indicated because you are using sample standard deviation.
critical t-score for 95% confidence interval with 8 degrees of freedom is equal to t = plus or minus 2.306004133.
t-score formula is t = (x - m) / s
t is the critical t-score
x is the critical raw score
m is the mean
s is the standard error.
standard error = standard deviation / sqrt(sample size) = 0.832165849 / sqrt(9) = .2773886163.
high side of the confidence interval t-score is 2.306004133 = (x - 98.366666667) / .2773886163.
solve for x to get x = 99.00632596.
low side of the confidence interval t-score is -2.306004133 = (x - 98.366666667) / .2773886163.
solve for x to get x = -2.306004133 * .2773886163 + 98.366666667 = 97.72700737.
your 95% confidence interval is from 97.72700737 to 99.00632596.
