Questions: You measure 20 randomly selected textbooks' weights, and find they have a mean weight of 53 ounces. Assume the population standard deviation is 13.5 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places 45.22 ≤ μ ≤ 60.78 The age when smokers first start from previous studies is normally distributed with a mean of 13 years old with a population standard deviation of 1.8 years old. A survey of smokers of this generation was done to estimate if the mean age has changed. The sample of 28 smokers found that their mean starting age was 12.6 years old. Find the 95% confidence interval of the mean. A simple random sample of n=20 is drawn from a population that is normally distributed with σ = 18. The sample mean is found to be x̄ = 62. Construct a 80% confidence interval for the population mean.

Solution Steps

To construct a 99% confidence interval for the true population mean textbook weight, we use the formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

• \(\bar{x} = 53\) (sample mean)
• \(\sigma = 13.5\) (population standard deviation)
• \(n = 20\) (sample size)
• \(z \approx 2.58\) (z-score for 99% confidence level)

Calculating the margin of error:

\[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} = 2.58 \cdot \frac{13.5}{\sqrt{20}} \approx 7.78 \]

Thus, the confidence interval is:

\[ (53 - 7.78, 53 + 7.78) = (45.22, 60.78) \]

To find the 95% confidence interval for the mean starting age of smokers, we apply the same formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

• \(\bar{x} = 12.6\) (sample mean)
• \(\sigma = 1.8\) (population standard deviation)
• \(n = 28\) (sample size)
• \(z \approx 1.96\) (z-score for 95% confidence level)

Calculating the margin of error:

\[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} = 1.96 \cdot \frac{1.8}{\sqrt{28}} \approx 0.67 \]

Thus, the confidence interval is:

\[ (12.6 - 0.67, 12.6 + 0.67) = (11.93, 13.27) \]

For the 80% confidence interval for the population mean, we again use the same formula:

\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]

Where:

• \(\bar{x} = 62\) (sample mean)
• \(\sigma = 18\) (population standard deviation)
• \(n = 20\) (sample size)
• \(z \approx 1.28\) (z-score for 80% confidence level)

Calculating the margin of error:

\[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} = 1.28 \cdot \frac{18}{\sqrt{20}} \approx 5.16 \]

Thus, the confidence interval is:

\[ (62 - 5.16, 62 + 5.16) = (56.84, 67.16) \]

Final Answer