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Vladimir [108]
3 years ago
8

Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are: 138.0 139.1 113.0 132.5 140.7 10

9.7 118.9 134.8 109.6 127.3 115.6 130.4 130.2 111.7 105.5 Assume that bushels per acre. a) Find the 90% confidence interval for the mean yield for this variety of corn. b) Find the 95% confidence interval. c) Find the 99% confidence interval. d) How do the margins of error in (a), (b), and (c) change as the confidence level increases?
Mathematics
1 answer:
son4ous [18]3 years ago
6 0
There is a relationship between confidence interval and standard deviation:
\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}
Where \overline{x} is the mean, \sigma is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first: 
\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}
When we do all the calculations we find that:
\overline{x}=123.8\\ \sigma=11.84
Now we can find confidence intervals:
($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing
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