Problem 34 - Basic Statistics for Business & Economics

Question

The human relations department of Electronics Inc. would like to include a dental plan as part of the benefits package. The question is: How much does a typical employee and his or her family spend per year on dental expenses? A sample of 45 employees reveals the mean amount spent last year was 1,820, with a standard deviation of 660. a. Construct a 95%$confidence interval for the population mean. b. The information from part (a) was given to the president of Electronics Inc. He indicated he could afford 1,700 of dental expenses per employee. Is it possible that the population mean could be 1,700 ? Justify your answer.

Step-by-Step Solution

Answer

Yes, $1,700 is within the 95% confidence interval, so it's possible the mean could be $1,700.

Step 1: Identify the given information

We have a sample of 45 employees with a mean annual dental expense of $1,820 and a standard deviation of $660. We need to construct a 95% confidence interval for the population mean.

Step 2: Determine the appropriate formula

We use the formula for a confidence interval for a population mean when the population standard deviation is unknown:\[\text{Confidence Interval} = \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}\]where \( \bar{x} \) is the sample mean, \( t_{\alpha/2} \) is the t-score, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Step 3: Find the critical t-score

To find \( t_{\alpha/2} \), we need the degrees of freedom, which is \( n-1 = 44 \). For a 95% confidence level, \( \alpha = 0.05 \), so \( \alpha/2 = 0.025 \). Using a t-table, we find the t-score for 44 degrees of freedom is approximately 2.015. This can vary slightly depending on which table or calculator is used.

Step 4: Calculate the margin of error

The margin of error is calculated as:\[E = t_{\alpha/2} \frac{s}{\sqrt{n}} = 2.015 \times \frac{660}{\sqrt{45}}\]Calculate this to find:\[E \approx 198.29\]

Step 5: Compute the confidence interval

The confidence interval is:\[(\bar{x} - E, \bar{x} + E) = (1820 - 198.29, 1820 + 198.29)\]Which simplifies to:\[(1621.71, 2018.29)\]

Step 6: Interpret the confidence interval

The 95% confidence interval for the population mean dental expenses is from $1,621.71 to $2,018.29. This means we are 95% confident that the true population mean falls within this range.

Step 7: Answer the president's question

The president can afford $1,700 per employee. Since $1,700 is within the confidence interval ($1,621.71 to $2,018.29), it is indeed possible that the population mean could be $1,700. However, this does not guarantee it will be exactly $1,700, it only indicates that $1,700 is a probable mean within the confidence range.

Key Concepts

Population Mean

The population mean is a statistical measure that represents the average of a given set of data for an entire population. In the context of the problem, it refers to the average amount spent on dental expenses by all employees and their families at Electronics Inc. Computing the population mean typically requires data from every member of the population, which is often impractical. Therefore, we rely on sample data to make an inference about the population mean. In this case, the sample mean of $1,820 from 45 employees is used to estimate the unknown true population mean.
This estimation is crucial because it helps in making decisions, like whether a dental plan is affordable, by providing a number to work with that likely represents the entire group.

Sample Standard Deviation

The sample standard deviation is an important statistic that measures the amount of variation or dispersion in a set of sample data. It quantifies how much the dental expenses vary from the sample mean of $1,820. In simpler terms, it shows if most employees spend around that average or if there is a wide range of expenses.
In our example, the sample standard deviation is $660. This value suggests that while some employees might spend significantly more or less than $1,820, this level of variation is normal for the sample size of 45 employees. Understanding this variation helps in calculating the confidence interval, which provides a range of likely values for the population mean.

T-Distribution

The t-distribution is utilized when the sample size is small, and the population standard deviation is unknown. It is similar to the normal distribution but accounts for uncertainty by being more spread out, especially with smaller samples.
In the context of the exercise, the t-distribution is critical because it provides the t-score necessary to determine the margin of error for the confidence interval. With 44 degrees of freedom (sample size minus one), we used a t-distribution to find a critical value of approximately 2.015 for a 95% confidence level. This t-score is essential to adjust for the sample size and to construct a reliable confidence interval for the population mean, knowing that using a normal distribution might underestimate the variability in small samples.

Margin of Error

The margin of error reflects the range of values above and below the sample mean where the true population mean is likely to fall. It recognizes the uncertainty inherent in estimating a population parameter from sample data.
In our example, we calculated the margin of error to be approximately $198.29. This figure is derived by multiplying the t-score (2.015) by the standard error of the mean, which is the sample standard deviation divided by the square root of the sample size. The calculated margin of error is then used to widen the estimate from the sample mean.
The result is a confidence interval that spans from $1,621.71 to $2,018.29. Thus, while the sample mean is $1,820, the true population mean could realistically be anywhere within this interval, including the $1,700 the president thinks the company can afford for dental expenses.