## Related questions with answers

Yvon Hopps ran an experiment to determine optimum power and time settings for microwave popcorn. His goal was to find a combination of power and time that would deliver high-quality popcorn with less than 10% of the kernels left un-popped, on average. After experimenting with several bags, he determined that power 9 at 4 minutes was the best combination. To be sure that the method was successful, he popped 8 more bags of popcorn (selected at random) at this setting. All were of high quality, with the following percentages of uncooked popcorn: 7, 13.2, 10, 6, 7.8, 2.8, 2.2, and 5.2. Does the 95% confidence interval suggest that he met his goal of an average of no more than 10% uncooked kernels? Explain.

Solution

VerifiedGiven:

$c=\text{Confidence level}=95\%=0.95$

DESCRIPTIVE MEASURES

The mean is the sum of all values divided by the number of values:

$\overline{x}=\dfrac{7+13.2+10+6+7.8+2.8+2.2+5.2}{8}=\dfrac{54.2}{8}=6.775$

$n$ is the number of values in the data set.

$n=8$

The variance is the sum of squared deviations from the mean divided by $n-1$. The standard deviation is the square root of the variance:

$s=\sqrt{\dfrac{(7-6.775)^2+....+(5.2-6.775)^2}{8-1}}\approx 3.6370$

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