Problem 39 - Basic Statistics for Business and Economics with Student

Question

A student conducted a study and reported that the 95 percent confidence interval for the mean ranged from 46 to \(54 .\) He was sure that the mean of the sample was 50 that the standard deviation of the sample was \(16,\) and that the sample was at least \(30,\) but could not remember the exact number. Can you help him out?

Step-by-Step Solution

Answer

The sample size is approximately 62.

Step 1: Understanding Confidence Interval

The 95% confidence interval given is from 46 to 54, which means the interval is centered around the sample mean. Since the student remembers the mean is 50, we verify that \[ \text{Confidence Interval} = \overline{x} \pm ME \]where ME is the margin of error. Here, \(\overline{x} = 50\).

Step 2: Calculate Margin of Error

The margin of error (ME) can be calculated using:\[ ME = 54 - 50 = 4 \text{ or } 50 - 46 = 4 \]Thus, the margin of error is 4.

Step 3: Formula for Margin of Error

For a 95% confidence interval and a large sample size (Normal Distribution assumption), the margin of error is given by:\[ ME = z \left( \frac{\sigma}{\sqrt{n}} \right) \]where \(z\) is the z-score for 95% confidence (approximately 1.96), \(\sigma\) is the standard deviation (16), and \(n\) is the sample size.

Step 4: Solve for Sample Size

Substitute the known values into the margin of error equation and solve for \(n\):\[4 = 1.96 \left( \frac{16}{\sqrt{n}} \right)\]Solve \(\sqrt{n} \approx \frac{16 \times 1.96}{4}\).Hence, \(\sqrt{n} \approx 7.84\) and squaring both sides gives \(n \approx 61.4664\).

Step 5: Conclude with Appropriate Sample Size

Since the sample size must be a whole number, round up to the nearest whole number to maintain statistical power. The sample size is \(n = 62\).

Key Concepts

Sample Size Calculation

To find an ideal sample size, we first use the margin of error and other known quantities, like the standard deviation, to reverse-engineer it. The sample size calculation begins when we notice that we're dealing with the formula for a confidence interval margin of error:

• Margin of Error (ME) = z-score \( \times \left(\frac{\sigma}{\sqrt{n}} \right) \)
• \(\sigma\) stands for standard deviation.
• \(n\) stands for sample size.

By knowing the margin of error and standard deviation, we solve for \(n\) by rearranging this formula. Consider that the z-score for a 95% confidence level is typically about 1.96. Through substitution and manipulation of the equation, we isolate \(n\) in order to compute it.

A step-by-step approach comprises substituting known values into the formula and proceeding to solve for \(\sqrt{n}\). Once solved, square \(\sqrt{n}\) to reveal \(n\). If \(n\) must be a whole number, round it up to ensure statistical reliability. This can be very significant in maintaining confidence in interpreting sample data.

Margin of Error

The Margin of Error (ME) is a critical component when estimating a population parameter. In simple terms, it tells us how much we can expect our sample mean to differ from the true population mean. For our exercise, the margin of error was given as 4.

To calculate the margin of error, the formula \(ME = z \left(\frac{\sigma}{\sqrt{n}}\right)\) is used:

• \(z\) is the z-score related to the desired confidence level.
• \(\sigma\) is the standard deviation of the sample.
• \(n\) is the sample size.

In our context, by knowing the confidence interval limits were 46 and 54 and center value (mean) is 50. We simply compute the margin to either boundary, either \(54 - 50\) or \(50 - 46\), both yielding 4. The margin of error essentially guides how "wide" the confidence interval will be, offering insight into the precision and reliability of the estimate.

Normal Distribution

When working with confidence intervals and sample mean estimates, the underlying data distribution often follows something called a Normal Distribution. This is important, as it informs us on why and how particular statistical rules hold.

A Normal Distribution is symmetric, characterized by its bell-shaped curve.

• The mean locates the center of the distribution.
• The standard deviation indicates how spread out the values are.

The assumption of normality allows statisticians to use z-scores to relate probabilities and intervals.

For practical use in statistics, when sample sizes are large (typically \(n \geq 30\)), the Central Limit Theorem assures us that the sampling distribution of the mean approaches normality. In turn, this justifies the use of what we call z-scores to calculate the confidence intervals, thus employing the properties of a normal distribution even if the actual data distribution isn't precisely "normal." This is a key benefit, as normal distribution models a range of phenomena in the natural and social sciences, making it versatile and widely applicable.