Question

PURELL Advanced Instant Hand Sanitizer kills more than 99.99% of most common germs. This alcohol-based hand sanitizer works in as little as 15 seconds, with no water or towels needed. Bottles of PURELL are supposed to contain, on average, 2 fluid ounces. A random sample of bottles of PURELL was drawn, and the weight of each was recorded (in fluid ounces). The following data was obtained: 2.16, 1.98, 2.04, 2.01, 1.93, 1.99, 2.04, 2.05, 1.96, 2.01. It is assumed that the weight of bottles of PURELL is normally distributed in the population. Construct a 99% confidence interval estimate of the true population mean weight of all bottles of PURELL Advanced Instant Hand Sanitizer. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Negative amounts should be indicated by a minus sign. Report your answer to 2 decimal places, using conventional rounding rules. Answer: D = [(Click to select) D o unces

Answer 1

From the data, we have,

$n=10,\overline{x}=\frac{\sum x}{n}=2.017$

$S=\sqrt{\frac{\sum (x_{i}-\overline{x})^2}{n-1}}=\sqrt{\frac{\sum x_{i}^2-n\overline{x}^2}{n-1}}=0.0629$

We know that the 99% confidence interval for population mean in case of a small sample is given by:

(T士 tar(0.01/2)。

$(2.017\pm t_{10-1}(0.01/2)\frac{0.0629}{\sqrt{10}})$

$(2.017\pm t_{9}(0.005)\frac{0.0629}{\sqrt{10}})$

0.0629 (2.017 (3.25) V10

$(2.017\pm 0.0646)$

So, the required interval is: $1.95\leq \mu\leq 2.08$