Question 1206353: In a survey, 13 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $35.6 and standard deviation of $3.3. Estimate how much a typical parent would spend on their child's birthday gift (use a 99% confidence level). Give your answers to 3 decimal places.
Found 2 solutions by Theo, math_tutor2020:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is 13.
sample mean is 35.6
standard deviation is 3.3
tk-score formula is t = (x-m)/s
t is the t-score
x is the critical sample mean
m is the sample mean
s is the standard error
at 99% two tail confidence interval with 12 degrees of freedom (sample size minus 1 = degrees of freedom in this problem), the critical t-score is equal to plus or minus t = 3.054540 rounded to 6 decimal places.
s = standard error = standard deviation / sqrt(sample size) = 3.3 / sqrt(13) = .915255 rounded to 6 decimal places.
t-score formula becomes plus or minus 3.054540 = (x - 35.6) / .915255.
on the high side of the confidence interval, solve for x to get x = 3.054540 * .915255 + 35.6 = 38.395683 rounded to 6 decimal places.
on the low side of the confidence interval, solve for x to get x = -3.054540 * .915255 = 35.5 = 32.804317 rounded to 6 decimal places.
your solution is that the a typical parent would spend from 32.804 to 38.396 on their child's birthday, rounded to 3 decimal places, at 99% confidence interval.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: (32.804, 38.396)
It is equivalent to writing 32.804 < mu < 38.396
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Explanation
n = 13 = sample size
xbar = 35.6 = sample mean
s = 3.3 = sample standard deviation
sigma = population standard deviation = unknown
Because we don't know the value of sigma, and because n > 30 is not the case, we must use the T distribution.
df = degrees of freedom
df = n-1
df = 13-1
df = 12
Use a T table such as this one
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
or one found on any other websites, or one found in the back of your stats textbook.
For exam purposes, your teacher is likely to hand out such a table if s/he expected you to use it.
Locate the row that starts with df = 12.
The column we want will have "confidence level = 99%" labeled at the bottom.
The intersection of this row and column leads to 3.055 which is the approximate t critical value.
It means that P(-3.055 < T < 3.055) = 0.99 approximately when df = 12.
Many online stats calculators can be used as an alternative to determine the t critical value.
Or you could use a TI84 or similar
https://www.statology.org/how-to-find-the-t-critical-value-on-a-ti-84-calculator/
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Here are the key values we need
xbar = 35.6
s = 3.3
n = 13
t = 3.055 (approximate)
Then,
E = margin of error
E = t*s/sqrt(n)
E = 3.055*3.3/sqrt(13)
E = 2.796105
which is approximate
Now we can compute the lower bound and upper bound.
L = lower bound
L = xbar - E
L = 35.6 - 2.796105
L = 32.803895
L = 32.804
U = upper bound
U = xbar + E
U = 35.6 + 2.796105
U = 38.396105
U = 38.396
The 99% confidence interval of the form L < mu < U would be roughly 32.804 < mu < 38.396
That condenses to the notation (32.804, 38.396)
Some textbooks will use the notation [32.804, 38.396] which represents the same thing more or less.
The fact we include endpoints or not isn't really important to the confidence interval overall.
Here is a calculator you can use to check your work
https://www.socscistatistics.com/confidenceinterval/default2.aspx
and here's another similar resource
https://www.statology.org/confidence-intervals-ti-84-calculator/
The function you want is called TInterval
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