Construct a 98% confidence interval for the population mean, $\mu.$ Assume the population has a normal distribution. A study of 14 bowlers showed that their average score was 192 with a standard deviation of 8.
A)(186.3, 197.7)
B)(222.3, 256.1)
C)(328.3, 386.9)
D)(115.4, 158.8)

Question

Quizplus · Accepted Answer

To construct a 98% confidence interval for the population mean, we use the formula: $\bar{x} \pm Z \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $Z$ is the Z-score corresponding to the confidence level. For a 98% confidence level, the Z-score is approximately 2.33 (from Z-tables or statistical software).Given: $\bar{x} = 192$, $s = 8$, $n = 14$.The margin of error (ME) is calculated as: $ME = Z \frac{s}{\sqrt{n}} = 2.33 \times \frac{8}{\sqrt{14}} \approx 5.0$.Thus, the confidence interval is: $192 \pm 5.0 = (187, 197)$.The correct choice is the one closest to our calculated interval, which is (186.3, 197.7).

Construct a 98% Confidence Interval for the Population Mean μ.\mu.μ. Assume the Population Has a Normal Distribution

Construct a 98% Confidence Interval for the Population Mean $\mu.$ Assume the Population Has a Normal Distribution