Question 1200758: In a survey, 30 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42.9 and a standard deviation of $8.3. 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.
Answer by math_tutor2020(3816)   (Show Source):
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Answer: 38.996 < mu < 46.804
The average amount spent is between $38.996 and $46.804
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Explanation:
At 99% confidence, the z critical value is roughly z = 2.576
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 99% confidence level.
xbar = 42.9 = sample mean
s = 8.3 = sample standard deviation
n = 30 = sample size
E = margin of error for the mean
E = z*s/sqrt(n)
E = 2.576*8.3/sqrt(30)
E = 3.90358215250216
E = 3.903582
L = lower boundary of confidence interval
L = xbar - E
L = 42.9 - 3.903582
L = 38.996418
L = 38.996
U = upper boundary of confidence interval
U = xbar + E
U = 42.9 + 3.903582
U = 46.803582
U = 46.804
The 99% confidence interval in the format (L,U) is approximately (38.996, 46.804) when rounding to 3 decimal places.
In the format L < mu < U, it would be 38.996 < mu < 46.804 which I think is a more descriptive way to write the confidence interval.
This is because we involve the parameter mu.
Side note: it's a bit strange that your teacher wants you to round to 3 decimal places, when usually we should round dollar figures to the nearest cent.
I would ask for clarification.
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