Question

You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately σ=68.7 dollars. You would like to be 98% confident that your estimate is within 1 dollar(s) of average spending on the birthday parties. How many moms do you have to sample? Do not round mid-calculation.n = 25535

Answer 1

Solution

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given parameters

$\begin{gathered} \sigma=68.7 \\ C.I=98\% \\ n=? \end{gathered}$

STEP 2: Get the z-score for the given Confidence interval

98% confidence is equivalent to 2.33 standard deviations. So you want $1 = 2.33 σ. That tells us the standard deviation of the sample mean needs to be:

$\frac{1}{2.33}=0.429184549$

The standard deviation of the sample mean is equal to:

$\frac{68.7}{\sqrt{n-1}}$

This implies that:

$\frac{68.7}{\sqrt{n-1}}=0.429184549$

STEP 3: Solve the equation for n

$\begin{gathered} \frac{68.7}{\sqrt{n-1}}=0.429184549 \\ \sqrt{n-1}=\frac{68.7}{0.429184549} \\ \\ \sqrt{n-1}=160.071 \end{gathered}$

Find the square of both sides:

$\begin{gathered} (\sqrt{n-1})^2=160.071^2 \\ n-1=25622.72504 \\ Add\text{ 1 to both sides} \\ n=25622.72504+1=25623.72504 \\ n\approx25624 \end{gathered}$