SOLUTION: In a survey, 12 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $33 and standard deviation of $10. Co

Question 1201712: In a survey, 12 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $33 and standard deviation of $10. Construct a confidence interval at a 98% confidence level.
Give your answers to one decimal place.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.

TWIN problems were solved many times at this forum.
Below is a short list of solutions (probably, not full list)

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1189932.html

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1079027.html

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1175460.html

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1109598.html

Happy learning (!)

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Answer: (25.2, 40.8)

========================================================================
Work Shown

Parameters
mu = population mean = unknown
sigma = population standard deviation = unknown

Statistics
n = 12 = sample size
xbar = 33 = sample mean
s = 10 = sample standard deviation

The goal is to estimate mu with a confidence interval.

Because sigma is not known, and because n > 30 is not the case, we must use the T distribution.

n = 12
df = degrees of freedom
df = n-1
df = 12-1
df = 11

Use a table such as this one
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to locate the df = 11 row and 98% confidence interval column.
The confidence level labels are listed at the bottom.

The value in this row and column is roughly 2.718 which is the critical t value.

We have these input values
t = 2.718
s = 10
n = 12

Calculate the margin of error
E = t*s/sqrt(n)
E = 2.718*10/sqrt(12)
E = 7.84619015828701
E = 7.846190

Then,
L = lower boundary of confidence interval
L = xbar - E
L = 33 - 7.846190
L = 25.15381
L = 25.2
and
U = upper boundary of confidence interval
U = xbar + E
U = 33 + 7.846190
U = 40.84619
U = 40.8

The 98% confidence interval in the format
L < mu < U
is approximately
25.2 < mu < 40.8

That condenses to
(25.2, 40.8)
I prefer the 1st format mentioned because it tells us what parameter we're trying to estimate.
But most textbooks and other settings will use the shortened 2nd format more often.

Interpretation: We are 98% confident the population mean amount spent on their child's birthday gift is somewhere between $25.20 and $40.80".

RELATED QUESTIONS
In a survey, 30 people were asked how much they spent on their child's last birthday... (answered by math_tutor2020)
In a survey, 13 people were asked how much they spent on their child's last birthday... (answered by Theo,math_tutor2020)
In a survey, 11 people were asked how much they spent on their child's last birthday... (answered by ewatrrr)
In a survey, 29 people were asked how much they spent on their child's last birthday... (answered by Boreal)
In a survey, 21 people were asked how much they spent on their child's last birthday... (answered by math_tutor2020)
In a survey, 30 people were asked how much they spent on their child's last birthday... (answered by Boreal)
In a survey, 30 people were asked how much they spent on their child's last birthday... (answered by Boreal)
In a survey, 29 people were asked how much they spent on their child's last birthday... (answered by Boreal)
In a survey, 11 people were asked how much they spent on their child's last birthday... (answered by Theo)