Problem 55 - Basic Statistics for Business & Economics

Question

A survey of 36 randomly selected iPhone owners showed that the purchase price has a mean of \(\$ 650\) with a sample standard deviation of 24 . a. Compute the standard error of the sample mean. b. Compute the 95% confidence interval for the mean. c. How large a sample is needed to estimate the population mean within 10 ?

Step-by-Step Solution

Answer

SE = 4; CI = (642.16, 657.84); Sample size = 23.

Step 1: Calculate the Standard Error of the Sample Mean

The standard error (SE) of the sample mean is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} \] where \( s = 24 \) is the sample standard deviation and \( n = 36 \) is the sample size. Plug in these values: \[ SE = \frac{24}{\sqrt{36}} = \frac{24}{6} = 4 \] Therefore, the standard error of the sample mean is 4.

Step 2: Calculate the 95% Confidence Interval for the Mean

The 95% confidence interval (CI) is given by the formula: \[ \text{CI} = \bar{x} \pm z \times SE \] where \( \bar{x} = 650 \) is the sample mean, and \( z \approx 1.96 \) for a 95% confidence level. We've previously calculated the standard error as 4. Now compute the confidence interval: \[ \text{CI} = 650 \pm 1.96 \times 4 \] Calculate the margin of error: \[ 1.96 \times 4 = 7.84 \] Then the confidence interval ranges from: \[ 650 - 7.84 = 642.16 \] to \[ 650 + 7.84 = 657.84 \] So, the 95% confidence interval for the mean is (642.16, 657.84).

Step 3: Calculate Sample Size Needed for Desired Margin of Error

To find the sample size needed to estimate the population mean within a margin of error \( E = 10 \), use the formula: \[ n = \left( \frac{z \times s}{E} \right)^2 \] Plug in the values where \( z = 1.96 \), \( s = 24 \), and \( E = 10 \): \[ n = \left( \frac{1.96 \times 24}{10} \right)^2 \] Calculate the expression: \[ \frac{1.96 \times 24}{10} = 4.704 \] Square the result: \[ n = 4.704^2 \approx 22.13 \] Since we cannot have a fraction of a subject, we round up to the nearest whole number: \( n = 23 \). Thus, a sample size of 23 is needed.

Key Concepts

Standard Error

The Standard Error (SE) is a statistical term that refers to the estimated standard deviation of a sample mean. It's a measure that helps us gauge how much the sample mean would differ from the actual population mean. Let's break this down for clarity:

Consider the formula for calculating Standard Error: \[ SE = \frac{s}{\sqrt{n}} \]Where:

• \( s \) is the sample standard deviation, which in our exercise is \( 24 \).
• \( n \) is the sample size, in this case, \( 36 \).

The standard error can be calculated by plugging these values into the formula, yielding \( SE = \frac{24}{6} = 4 \).

In simpler terms, the standard error gives us an idea of how much we can expect the sample mean to "wiggle" around the actual population mean. The smaller the standard error, the more accurate our sample mean is as an estimate of the population mean.

Sample Size Calculation

Sample size plays a crucial role in determining how precisely you can estimate the population mean. Essentially, the larger your sample size, the more reliable your estimate will be.

For our exercise, we needed to find out how many sample subjects are required to estimate the population mean within a certain margin of error (\( E \)). This is achieved using the formula:\[ n = \left( \frac{z \times s}{E} \right)^2 \]Where:

• \( z \) is the z-score (1.96 for a confidence level of 95%).
• \( s \) is the sample standard deviation.
• \( E \) is the desired margin of error, set at \( 10 \) in this case.

By substituting the known values into the formula, \[ n = \left( \frac{1.96 \times 24}{10} \right)^2 = 22.13 \]We round up to \( 23 \) because sample sizes must be whole numbers. As you can see, having the right sample size ensures that your estimates will be statistically sound.

Margin of Error

The Margin of Error is the amount that you allow between your sample statistic and the actual population parameter. It’s a way of expressing the range within which you can confidently claim the actual value lies.

For a 95% confidence interval calculation, which we conducted in this exercise, the margin of error is calculated by multiplying the z-score by the standard error. The formula looks like this:\[ ME = z \times SE \]Using our specific exercise values:

• \( z \) is 1.96, representing our 95% confidence level.
• \( SE \) is 4.

Thus:\[ ME = 1.96 \times 4 = 7.84 \]This indicates that the actual population mean could differ by as much as \( 7.84 \) from the sample mean. Knowing the margin of error helps to provide a range, such as the confidence interval calculated, where the true population mean is likely to reside, thereby offering a clearer picture of the uncertainty associated with your findings.