Question

Mr. Bond’s class wanted to estimate the mean mass of Snickers Fun Size bars. They randomly selected 74 bars and recorded the mass of each unwrapped bar. The bars had a mean mass of 17.1g with a standard deviation of 0.76g. Construct and interpret a 99% confidence interval for µ = the true mean mass of Snickers Fun Size bars.

Answer 1

Step 1: Find the standard error (SE)

The standard error is given by

$SE=\frac{s}{\sqrt[]{n}}$$\begin{gathered} \text{ Where } \\ SE=\text{ the standard error} \\ s=\text{ the sample standard deviation} \\ n=\text{ the sample size} \end{gathered}$

In this case,

$n=74,s=0.76$

Therefore,

$SE=\frac{0.76}{\sqrt[]{74}}\approx0.0883$

Step 2: Find the alpha level (α)

$\alpha=1-\frac{(\text{Confidence level})}{100}$$\alpha=1-0.99=0.01$

Step 3: Find the critical probability (P*)

$P^{\prime}=1-\frac{\alpha}{2}$

Therefore,

$P^{\cdot}=1-\frac{0.01}{2}=0.995$

Step 4: Find the critical value (CV)

The critical value the z-score having a cumulative probability equal to the critical probability (P*).

Using the cumulative z-score table we will find the z-score with value of 0.995

Hence,

$CV=2.576$

Step 5: Find the margin of error (ME)

$ME=SE\times CV$

Therefore,

$ME=0.0883\times2.576=0.2275$

Step 6: Find the confidence interval (CI)

$\begin{gathered} CI\text{ is given by} \\ CI=(\bar{x}-ME,\bar{x}+ME) \\ \text{ In this case} \\ \bar{x}=17.1 \end{gathered}$

Therefore,

$CI=(17.1-0.2275,17.1+0.2275)=(16.8725,17.3275)$

Hence there is a 99% probability that the true mean will lie in the confidence interval

(16.8725, 17.3275)