Problem

MATH1127: Introduction to Statistics (60022) Lesson 8.2 A Single Population Mean using the Student
Unit 3 Chapter 8: Lesson 8.2 Assignment
Score: $87 / 100 \quad 17 / 20$ answered
Question 12
In a survey, 14 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $\$ 31$ and standard deviation of $\$ 14$. Construct a confidence interval at a $99 \%$ confidence level.
Give your answers to one decimal place.
$\pm$

Answer

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Answer

Subtracting and adding this margin of error from the sample mean, we find that the 99% confidence interval for the population mean is \(\boxed{(\$19.7, \$42.3)}\).

Steps

Step 1 :We are given that the sample mean (\(\bar{x}\)) is $31, the sample standard deviation (s) is $14, and the sample size (n) is 14. We want to construct a confidence interval at a 99% confidence level.

Step 2 :The formula for a confidence interval for a population mean when the population standard deviation is unknown is \(\bar{x} \pm t \frac{s}{\sqrt{n}}\), where t is the t-score corresponding to the desired confidence level and degrees of freedom (which is n - 1).

Step 3 :We need to find the t-score for a 99% confidence level and 13 degrees of freedom. Using a t-distribution table or a calculator, we find that the t-score is approximately 3.012.

Step 4 :Substituting the given values into the formula, we get \(31 \pm 3.012 \frac{14}{\sqrt{14}}\).

Step 5 :Solving this expression, we find that the margin of error is approximately 11.27.

Step 6 :Subtracting and adding this margin of error from the sample mean, we find that the 99% confidence interval for the population mean is \(\boxed{(\$19.7, \$42.3)}\).

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