Question

Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are:

138.0 139.1 113.0 132.5 140.7 109.7 118.9 134.8
109.6 127.3 115.6 130.4 130.2 111.7 105.5
Assume that Math Notation 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?

Answer 1

@ParthKohli

Answer 2

@jim_thompson5910

Answer 3

@jigglypuff314

Answer 4

Are you familiar with the confidence interval formula?

Answer 5

no @jim_thompson5910

Answer 6

so you've never seen these formulas before? or anything like it?

\[\Large L = \bar{x} - z^{*} \frac{s}{\sqrt{n}}\]
\[\Large U = \bar{x} + z^{*} \frac{s}{\sqrt{n}}\]

Answer 7

no :/ flvs kinda sucks @jim_thompson5910

Answer 8

ok let's go over the formulas then. They aren't too bad

the first symbol that we'll go over is this

\[\LARGE \bar{x}\]

that's pronounced "x-bar" and it is the sample mean

to find the sample mean, you will


step 1) add up the values
step 2) divide the result (from step 1) by the number of values

tell me what you get

Answer 9

123.8 @jim_thompson5910

Answer 10

very good. Ok onto the next symbol \(\LARGE z^{*}\)
which is often pronounced "z-star".

This is the critical z-value based on the confidence level

It is impossible to calculate this by hand (well it is possible but it will take a very long time). So we can use a calculator or a table

Let's use the table

https://www.ltcconline.net/greenl/courses/201/estimation/smallConfLevelTable.htm

when the confidence level is 0.90, what is the value of z?

Answer 11

1.645? @jim_thompson5910

Answer 12

yes, so \(\LARGE z^{*} = 1.645\) for the 90% confidence level

Answer 13

the 's' in the formula refers to the sample standard deviation. Let's use a calculator to compute this

if you don't have a calc that can handle standard deviation, then use this one

http://www.miniwebtool.com/sample-standard-deviation-calculator/

Answer 14

so is that my a? the 1.645? @jim_thompson5910

Answer 15

no, we're still working on the pieces of the formula

Answer 16

part a) isn't done yet

Answer 17

here's how you can enter the data (see attached)

Answer 18

and for the sample standard deviation i got 12.2603891805 @jim_thompson5910

Answer 19

me too

Answer 20

ok so
s = 12.260389
approximately

Answer 21

finally the n is simply the sample size. No need to use a table or calc here. We just look at the problem

it says `Crop researchers plant 15 plots ` so n = 15

Answer 22

let's plug these values in to get...


\[\Large L = \bar{x} - z^{*} \frac{s}{\sqrt{n}}\]
\[\Large L = 123.8 - 1.645* \frac{12.260389}{\sqrt{15}}\]
\[\Large L = ???\]


Use a calculator to compute the expression and tell me what you get

Answer 23

i received 386.6961679 @jim_thompson5910

Answer 24

that's incorrect. How did you enter that into your calc?

Answer 25

just how you wrote it, but let me try again @jim_thompson5910

Answer 26

i got 118.592557 this time idk what happened the last time lol @jim_thompson5910

Answer 27

you can try this calc

http://web2.0calc.com/

and type in `123.8 - 1.645*(12.260389/sqrt(15))`

Answer 28

much better, yes L = 118.592557 approx

Answer 29

this is the lower bound of the confidence level
L = Lower bound

U is the upper bound

so let's calculate that now

\[\Large U = \bar{x} + z^{*} \frac{s}{\sqrt{n}}\]
\[\Large U = 123.8 + 1.645* \frac{12.260389}{\sqrt{15}}\]
\[\Large U = ???\]

Answer 30

is the upper one 129.007443? @jim_thompson5910

Answer 31

yep

Answer 32

so the 90% confidence interval stretches from 118.592557 to 129.007443

Answer 33

is that a? @jim_thompson5910

Answer 34

let's round each value to 1 decimal place to get 118.6 and 129.0

so the 90% confidence interval is (118.6, 129.0)

agreed?

Answer 35

I'm rounding to 1 decimal place because the other data values are given to 1 decimal place. Your book or teacher may want you to round to 2 decimal places

Answer 36

okay :)

Answer 37

yes once you get to (118.6, 129.0) or (118.59, 129.01), then you're done with part a)

Answer 38

yay!! now b lol @jim_thompson5910

Answer 39

part b) is going to be the same as part a) but the only difference is that z value will change

use the table and tell me what z is equal to when the confidence level is 95%

Answer 40

1.96 @jim_thompson5910

Answer 41

good

Answer 42

repeat part a) but just use z = 1.96 instead

Answer 43

xbar, s and n will remain the same

Answer 44

for te U i got 130.0046129 @jim_thompson5910

Answer 45

good

how about L?

Answer 46

117.5953871 ? @jim_thompson5910

Answer 47

good

Answer 48

if we round to 1 decimal place, the CI (confidence interval) is (117.6, 130.0)

if we round to 2 decimal places, the CI is (117.60, 130.00)

so effectively the two results are identical

Answer 49

yayy @jim_thompson5910

Answer 50

for part c) you'll do the same as a) & b) but as you can probably guess by now, the z value will change.

xbar, s, & n will stay the same

Answer 51

sweet this is so much easier than i thought it would be thank you !! @jim_thompson5910

Answer 52

then in part d), you just make an observation of what is happening (as the confidence level increases, do the intervals increase in size? or do the intervals decrease in size? or do the intervals stay the same?)

Answer 53

or specifically you focus on the margin of error