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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.

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Answered 11 months ago
Answered 11 months ago
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Given:

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

DESCRIPTIVE MEASURES

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

x=7+13.2+10+6+7.8+2.8+2.2+5.28=54.28=6.775\overline{x}=\dfrac{7+13.2+10+6+7.8+2.8+2.2+5.2}{8}=\dfrac{54.2}{8}=6.775

nn is the number of values in the data set.

n=8n=8

The variance is the sum of squared deviations from the mean divided by n1n-1. The standard deviation is the square root of the variance:

s=(76.775)2+....+(5.26.775)2813.6370s=\sqrt{\dfrac{(7-6.775)^2+....+(5.2-6.775)^2}{8-1}}\approx 3.6370

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