A random sample of 25 fields of spring wheat has a mean yield of 27.6 bushels per acre and standard deviation of 5.69 bushels per acre. Determine the 98% confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2:Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 2:Construct the 98% confidence interval. Round your answer to one decimal place.

Step 1     z(c) = 2,323

CI = (  25 ;  30,2 )

Step-by-step explanation:

Normal Distribution:

Sample size  n = 25

Sample mean   μ  = 27,6

Sample standard deviation  s = 5,69

CI = 98 %   then  α  = 1 - 0,98     α = 0,02    α/2  = 0,01

From z-table we don´t  find z (score) for 0,01   directly, we need to interpolate between z = 2,32   and z = 2,33

For   0,0099           z score   is    2,33

for    0,0102            z score   is    2,32

Δ      0,0003                                  0,01

Then   by rule of three

for Δ        0,0003                ⇒      0,01

for Δ   (0,01- 0,0102)          ⇒        x

0,0003    0,01

0,0002      x

x = 0,0067

then  z (score for 0,01)   =  2,33 - 0,0067     z(c) = 2,3233

round to three decimal places z(c) = 2,323

Step 2:

CI =   μ  ±  z(c) * s/√n    ⇒   27,6  ± (2,323 * 5,69)/5

CI =  27,6 ± 2,6436

round to one decimal place

CI = 27,6 ± 2,6

CI = (  25 ;  30,2 )

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step-by-step explanation:

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this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

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