Questions: Use technology and the given confidence level and sample data to find the confidence interval for the population mean μ. Assume that the population does not exhibit a normal distribution. Weight lost on a diet: 99% confidence n=61 x̄=20 kg s=3.7 kg What is the confidence interval for the population mean μ ? Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed? A. No, because the population resembles a normal distribution. B. Yes, because the sample size is not large enough. C. No, because the sample size is large enough. D. Yes, because the population does not exhibit a normal distribution.

Solution Steps

We are provided with the following data for the weight lost on a diet:

• Confidence Level: $99\%$
• Sample Size: $n = 61$
• Sample Mean: $\bar{x} = 20 \, \text{kg}$
• Sample Standard Deviation: $s = 3.7 \, \text{kg}$

To find the confidence interval for the population mean $\mu$, we first calculate the margin of error using the formula:

$$\text{Margin of Error} = z \cdot \frac{s}{\sqrt{n}}$$

For a $99\%$ confidence level, the critical value $z$ is approximately $2.576$. Thus, we compute:

$$\text{Margin of Error} = 2.576 \cdot \frac{3.7}{\sqrt{61}} \approx 2.576 \cdot 0.473 \approx 1.220$$

The confidence interval for the population mean $\mu$ is given by:

$$\bar{x} \pm \text{Margin of Error}$$

Substituting the values, we have:

$$20 \pm 1.220$$

This results in the confidence interval:

$$(20 - 1.220, 20 + 1.220) = (18.780, 21.220)$$

Rounding to one decimal place, we find:

$$\text{Confidence Interval} = (18.8, 21.2)$$

Since the sample size $n = 61$ is greater than $30$, we can apply the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normally distributed regardless of the population's distribution. Therefore, the confidence interval is not significantly affected by the fact that the population does not exhibit a normal distribution.

Final Answer