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.

Question

contestada

Respuesta :

jtellezd jtellezd · Answer 1 · 2020-12-12T14:46:37+01:00

Answer:

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 )