Question

If you wish to estimate a population mean with a sampling distribution error SE = 0.28 using a 95% confidence interval and you know from prior sampling that o? is approximately equal to 4.1, how manyobservations would have to be included in your sample?The number of observations that would have to be included in your sample is(Round up to the nearest observation.)

Answer 1

Number of observations = 201

SOLUTION

Problem Statement

The question tells us to find the total number of observations needed to be included in our sample given that we have a standard error of 0.28 using a 95% confidence interval with a variance of 4.1.

Method

In order to solve this problem, we need to follow the steps listed below:

1. We need to find the alpha level for the sampling. The formula for the alpha level is given by:

`\begin{gathered} \alpha=1-\frac{\text{confidence level}}{\text{100}} \\ \\ \text{where,} \\ \alpha=\text{alpha level} \end{gathered}`

2. Find the critical probability (p*) from the alpha level gotten in step 1

`p=1-(\alpha)/(2)`

3. Convert the critical probability gotten above into a z-score (Z) by checking the z distribution table

4. Using the formula below, estimate the number of observations needed:

`\begin{gathered} \text{The initial formula is:} \\ ME=Z*\frac{\sigma}{\sqrt[]{n}} \\ \text{where,} \\ ME=\text{Margin of Error = Sampling distribution Error} \\ Z=\text{critical value gotten from z-table} \\ \sigma=\text{ standard deviation}=\text{square root of the variance} \\ n=\text{ number of observations} \\ \\ \text{Making n the subject of the formula, we have} \\ \therefore n=\mleft((Z*\sigma)/(ME)\mright)^2 \end{gathered}`

(Note: When the sampling is random, the maximum likely size of the sampling error is called the margin of error)

Implementation

1. We need to find the alpha level for the sampling:

`\begin{gathered} \text{confidence level = 95\% = }(95)/(100) \\ \alpha=1-(95)/(100)=1-0.95 \\ \therefore\alpha=0.05 \end{gathered}`

2. Find the critical probability (p*) from the alpha level gotten in step 1

`p=1-(0.05)/(2)=0.975`

3. Convert the critical probability gotten above into a z-score (Z) by checking the z distribution table:

Notice that the two loops (red and blue) intersect at the critical probability value of 0.9750 on the z-table. We read off the Critical value (Z) from the vertical axis at the left of the page first then the horizontal axis at the top of the page.

Thus, the Critical Value (Z) = 1.96

4. Using the formula below, estimate the number of observations needed:

`\begin{gathered} n=((Z*\sigma)/(ME))^2 \\ \sigma^2=4.1 \\ \therefore\sigma=\sqrt[]{4.1} \\ \\ n=\mleft(\frac{1.96*\sqrt[]{4.1}}{0.28}\mright)^2 \\ \\ n=200.9\approx201 \end{gathered}`

Final Answer

The number of observations is: 201