To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has
Recently been marketed. The FDA tests on this cigarette yielded mean nicotine content of 26.5
Milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes. Construct a
90% confidence interval for the mean nicotine content of this brand of cigarette.
A) $26.5 \pm 1.710$
B) $26.5 \pm 1.612$
C) $26.5 \pm 1.685$
D) $26.5 \pm 1.589$

Question

To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has
Recently been marketed. The FDA tests on this cigarette yielded mean nicotine content of 26.5
Milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes. Construct a
90% confidence interval for the mean nicotine content of this brand of cigarette. 
A) $26.5 \pm 1.710$
B) $26.5 \pm 1.612$
C) $26.5 \pm 1.685$
D) $26.5 \pm 1.589$

Quizplus · Accepted Answer

To construct a 90% confidence interval for the mean nicotine content, we use the formula: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score corresponding to the confidence level and degrees of freedom (df = n - 1). For a 90% confidence interval and df = 8, the t-score (from a t-table or calculator) is approximately 1.860. Plugging in the values: $26.5 \pm 1.860 \times \frac{2.6}{\sqrt{9}} = 26.5 \pm 1.612$.

To Help Consumers Assess the Risks They Are Taking, the Food