Math211-T/Th-I Homework: Week 4 Homework Score: 0.67 of 2 pts 7.1.18 experiment

Question

Math211-T/Th-I Homework: Week 4 Homework Score: 0.67 of 2 pts 7.1.18 experiment with peas resulted in one sample of offspring that consisted of 434 green peas and 160 yellow peas A genetic a. Construct a 95% confidence interval to estimate of the percentage of yellow peas b. It was expected that 25% of the offspring peas would be yellow Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? a. Construct a 95% confidence interval Express the percentages in decimal form p(Round to three decimal places as needed) swer in the edit fields and then click Check Answer

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
possibile chances (x)=160
sample size(n)=434
success rate ( p )= x/n = 0.3687
I.
sample proportion = 0.3687
standard error = Sqrt ( (0.3687*0.6313) /434) )
= 0.0232
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0232
= 0.0454
III.
CI = [ p ± margin of error ]
confidence interval = [0.3687 ± 0.0454]
= [ 0.3233 , 0.4141]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=160
sample size(n)=434
success rate ( p )= x/n = 0.3687
CI = confidence interval
confidence interval = [ 0.3687 ± 1.96 * Sqrt ( (0.3687*0.6313) /434) ) ]
= [0.3687 - 1.96 * Sqrt ( (0.3687*0.6313) /434) , 0.3687 + 1.96 * Sqrt ( (0.3687*0.6313) /434) ]
= [0.3233 , 0.4141]

PART B.
it is found that the percentage of expected yellow peas are b/w 32% to 41%, since it is said that the percentage
of yellow peas is not 25% , we are agreed to the statement

MINITAB RESULT

Test and CI for One Proportion

Sample X N Sample p 95% CI
1 160 434 0.368664 (0.323154, 0.415977)

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---