SOLUTION: ESTIMATING THE POPULATION MEAN WITH UNKNOWN VARIANCE USING A SMALL SAMPLE The mean length 𝜇 of time consumed in online classes per day of Senior High School students would ha

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Question 1182597: ESTIMATING THE POPULATION MEAN WITH UNKNOWN VARIANCE USING A SMALL SAMPLE
The mean length 𝜇 of time consumed in online classes per day of Senior High School students would have to be estimated. A sample of 20 students had a mean of 3.5 hours a day with a sample standard deviation of 0.8 hour. Find the 95 % confidence interval for the 𝜇.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the sample mean is equal to 3.5 hours.
the sample standard deviation is equal to .8 hours.
the sample size is 20.

t-score would be used with 19 degrees of freedom is what i think.
at 95% confidence level, the two tail critical t-score with 19 degrees of freedom would be equal to plus or minus 2.093.

19 degrees of freedom is because the sample size is 20.

the standard error would be equal to .8 / sqrt(20) = .1789.
that's the standard deviation of the sample divided by the square root of the sample size.

the t-score formula becomes plus or minus 2.093 = (x - 3.5) / .1789

solve for x to get:

x = .1789 * 2.093 + 3.5 on the high side and x = .1789 * -2.093 + 3.5 on the low side.

x will be equal to 3.8744377 on the high side.
x will be equal to 3.1255623 on the low side.

your confidence interval is from 3.1256 on the low side to 3.8744 on the high side when the solution is rounded to 4 decimal places.

the graph of the critical t-score would look like this:

you need to find the standard error and use the t-score formula to find the raw scores from the t-scores.

i did that above.

standard error = standard deviation of sample divided by square root of sample size.

t-score formula is t = (x = m) / s
t is the t-score
x is the raw score
m is the mean
s is the standard error