The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 27 kg with a sample standard deviation of 9 kg. Assume a normal population. Develop a 90% confidence interval for the mean annual consumption of sugar.
A) 23.06 to 30.94
B) 22.20 to 31.80
C) 20.37 to 33.63
D) 23.30 to 30.70
E) 23.0 to 30.0

Question

Quizplus · Accepted Answer

To calculate the 90% confidence interval for the mean, we use the formula: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score for 90% confidence and 15 degrees of freedom (since n-1 = 16-1 = 15), $s$ is the sample standard deviation, and $n$ is the sample size. Given $\bar{x} = 27$, $s = 9$, and $n = 16$, and knowing the t-score for 90% confidence and 15 df is approximately 1.753 (from t-tables), the interval is calculated as $27 \pm 1.753 \times \frac{9}{\sqrt{16}} = 27 \pm 3.94$, which gives us 23.06 to 30.94.

The Sugar Producers Association Wants to Estimate the Mean Yearly