SOLUTION: A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people
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Question 1193902: A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below. From past studies, the publisher assumes that the population of times is normally distributed. Construct the 95% confidence interval for the population mean.
12 8 11 10 11 6 6 9 6 11 11 11 7 7 9
The sample mean is
The sample standard deviation rounded to 2 decimal places is
The number of degrees of freedom is
According to the t-distribution table in the formula sheets t c rounded to 3 decimal places is
The margin of error E rounded to two decimal places is
The confidence interval is < 𝜇 <
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
There are n = 15 values in the list, which is the sample size.
To get the sample mean, first we add up the values
12+8+11+10+11+6+6+9+6+11+11+11+7+7+9 = 135
Then divide this over the sample size (n = 15) to get
135/15 = 9
The sample mean is 9
We call this xbar because x has a horizontal bar over top.
xbar = 9
You could calculate the sample standard deviation by hand, but it's preferable to use technology instead. There are many free calculators online to help with that if you don't have a TI83/84.
You should get a sample standard deviation of roughly s = 2.17
The sample standard deviation s estimates the population standard deviation sigma.
s = sample standard deviation = 2.17 approximately
sigma = population standard deviation = unknown
Because we don't know sigma and because n > 30 is not true, we must use a T distribution.
The degrees of freedom (df) is equal to the sample size minus 1
df = n-1
df = 15-1
df = 14
Use the T table in the back of your textbook, or use a free online resource like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Start by looking at the row labeled "two tails". The value 0.05 corresponds to a confidence interval of 95% since 1 - 0.05 = 0.95
The 0.05 refers to the combined area of both tails.
Highlight this entire column.
Then look in the row that has df = 14 at the very left.
The intersection of the row and column leads to the approximate t critical value of 2.145
What is this value saying? It says that P(-2.145 < T < 2.145) = 0.95 approximately when df = 14.
95% of the area under the T curve (df = 14) is between roughly t = -2.145 and t = 2.145
Let's compute the margin of error
And lastly, let's compute the confidence interval for the mean mu (symbol 𝜇)
We could shorten this to (7.80, 10.20) which is common notation for confidence intervals.
Some books and research papers use the notation [7.80, 10.20] to mean the same thing.
We are 95% confident the true population mean (mu) is somewhere between 7.80 and 10.20
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Answers:
Sample mean = 9
Sample standard deviation = 2.17
Degrees of Freedom = 14
(t critical value) is roughly 2.145
Margin of error: E = 1.20
The 95% confidence interval is roughly
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