Question

Suppose you want to estimate one of four parameters- \(\mu, \mu_{1}-\mu_{2}, p,\) or \(p_{1}-p_{2}-\) to within a given bound with a certain amount of confidence. Use the information given to find the appropriate sample size(s). Estimating \(\mu_{1}-\mu_{2}\) to within .17 with probability. \(90 .\) Assume that the sample sizes will be equal and that \(\sigma_{1}^{2} \approx \sigma_{2}^{2} \approx 27.8\).

Short answer

Answer: 259

Step-by-step solution

Identify the critical value corresponding to the confidence level

Since we want a 90% confidence level, we need to find the critical value, \(Z_{\alpha/2}\), which corresponds to the area in the tail of the standard normal distribution. Using a standard normal distribution table or a calculator, we find \(Z_{\alpha/2} \approx 1.645\).

Use the given bound, population variances and critical value to find the required sample size

We are given the following information: - The desired bound for the estimation of \(\mu_{1}-\mu_{2}\) is 0.17. - The population variances are approximately equal, with \(\sigma_{1}^{2} \approx \sigma_{2}^{2} \approx 27.8\). - The sample sizes will be equal, \(n_{1} = n_{2} = n\). We can write the margin of error formula for the difference between two population means as follows: $$E = Z_{\alpha/2} \cdot \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}$$ Given the problem's constraints, we can simplify this formula: $$E = Z_{\alpha/2} \cdot \sqrt{\frac{\sigma^{2}}{n} + \frac{\sigma^{2}}{n}}$$ Now plug in the given values: $$0.17 = 1.645 \cdot \sqrt{\frac{27.8}{n} + \frac{27.8}{n}}$$

Solve for the required sample size, \(n\)

First, we will square both sides of the equation to eliminate the square root: $$(0.17)^{2} = (1.645)^{2} \cdot \left(\frac{27.8}{n} + \frac{27.8}{n}\right)$$ Now simplify the equation: $$0.0289 = 2.706 \cdot \frac{27.8}{n}$$ Next, isolate the \(n\) term: $$n = \frac{2.706 \cdot 27.8}{0.0289}$$ We get: $$n \approx 258.59$$

Round up the sample size

Since we cannot have a fraction of a sample, we round up the sample size to the nearest whole number: $$n = 259$$ Thus, to estimate the difference between the two population means to within a bound of 0.17 with a 90% confidence level, we require a sample size of 259 for each population.

Key concepts

Confidence Interval

When we talk about estimating a parameter like a population mean difference, a confidence interval is a crucial concept. It gives us a range within which we can say with a certain degree of certainty that the true parameter value lies. The probability associated with this degree of certainty is known as the confidence level. For example, a 90% confidence level means that, were we to take many random samples and calculate the interval each time, about 90% of these intervals would contain the true population mean difference.

In the provided exercise, to find the appropriate sample size for estimating the population mean difference within a specific margin of error (0.17) with a 90% certainty, we needed to identify the critical value from the standard normal distribution. This value helps define the width of the confidence interval. The margin of error is also influenced by the population variances and the sample size. Thus, determining an adequate sample size ensures the confidence interval is neither too wide nor too narrow, reflecting an effective balance between precision and resource expenditure.

Population Mean Difference

The population mean difference, \frac{\(mu_1 - \(mu_2}}{\text{Population Mean Difference}}}\), represents the difference between the means of two independent populations. In research or practical applications, comparing two means is often necessary to understand whether there exists a significant difference between two groups. This could be the efficacy of two different drugs or teaching methods, for example.

In our exercise, the goal was to estimate this difference to a specific precision, expressed as the bound within the confidence interval. To achieve the desired precision, both the sample and population variances are factored into the calculation. When the variances are assumed to be equal and the sample sizes for both groups are the same, the formula simplifies, making it easier to calculate the required sample size. The understanding of population mean difference is important for accurate estimation and has direct implications on the sample size calculation.

Standard Normal Distribution

The standard normal distribution, often symbolized as Z, is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. This distribution is fundamental in statistics because we can use it to determine probabilities and critical values for normal distributions with different means and standard deviations.

For example, in the exercise, the critical value needed for the 90% confidence interval was found using the standard normal distribution. This critical value (1.645) reflects the point in the distribution where the area under the curve to the right (or left for a two-tailed test) equals 5% (since 90% confidence means we are leaving 10% in the tails, split into two halves). It serves as a multiplier to adjust the margin of error for the desired level of confidence. The standard normal distribution is a key player in sample size determination and confidence intervals because it provides a universal reference for calculating probabilities associated with the normal random variables.