A genetic experiment with peas resulted in one sample of offspring that consisted of 447 green peas and 151 yellow peas.

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) Answer: The 95% confidence interval of population proportion(p) is (0.2519,0.2531).

b) Conclusion: There is not sufficient evidence to conclude that the percentage of offspring yellow peas is not? 25%.
Yes,the results contradict? expectations.

Step-by-step explanation

Question #a:
Let's write the given information:
Here variable for which have to find confidence interval is yellow peas
n = sample size =447+151=598 , x =number of yellow peas=151 
 

The point estimate of the population proportion is sample proportion p? = nx?=598151?=0.2525.
 

The formula of confidence interval for population proportion(p) is as follow:
 

(LowerLimit,UpperLimit)=(p?−E,p?+E......................... ...(1)
 

The formula of margin of error(E) to estimate confidence interval for population proportion is as follows:
 

E=Zc?∗np?∗(1−p?)??   ................(2)
 

Let's find Zc
It is given that ; c = confidence level = 0.95 
So that  level of significance = α = 1 - c = 1 - 0.95 = 0.05
this implies that α/2 = 0.05/2 = 0.025
So we want to find   Zc  such that  
P(Z > Zc) = 0.0250.
Therefore , P(Z < Zc) = 1 - 0.025 = 0.9750
From z-table, the z-score corresponding to the probability 0.9750 is 1.96.
Note: Using excel, Zc = "=NORMSINV(0.975)" = 1.96
So for n = sample size =598 , \widehat p =0.2525 , and Zc = 1.96, we get 
 

Plugging these values in the formula of E, we get,
 

E=1.96∗5980.2525∗(1−0.2525)??

E= 1.96* 0.000316=0.000619
So we get Margin of Error, E = 0.000619

Lowerlimit=p?−E=0.2525−0.000619=0.2519
Upperlimit=p?+E=0.2525+0.000619=0.2531
Answer: The 95% confidence interval of population proportion(p) is (0.2519,0.2531).

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b)

Claim: The proportion of yellow peas is different than 25%=0.25 
From the above claim, the null hypothesis(H0) and the alternative hypothesis(H1) are as follows:
H0: p =0.25 
H1: p ≠ 0.25
Since the alternative hypothesis include "≠" inequality, so it is two tailed test.
 

Here we need to use one sample proportion z-test.
 

Let's write the given information: 
n = sample size =598,
x = number of yellow peas=151
Sampleproportion=p?=nx?

p?=598151?=0.2525

In the b) part level of significance (α) is not given so we will take it as alpha= 0.05
and p =0.25  ...(from the null hypothesis)
 

Let's find the value of test statistic:
The formula of z-test statistic for one sample proportion z-test is as follows:
z=np∗(1−p)??p?−p?.

Plugging the given information in the above formula, we get,

z=5980.25∗(1−0.25)??0.2525−0.25?=0.01770.0025?=0.1412

∴ The test statistic value = z =0.1412=0.14.......(Round it to two decimal places)
 

Let's find P-value:
For two tailed z-test with positive z value, the P-value is = 2*P(Z < -z) 
We need to use technology as "Excel" to find the p-value.
The gereal command to find the less than standard normal probability in excel is  "=NORMSDIST(-z)"
Here z =0.14 
?∴ 2*P(Z <-0.14 ) = "=2*NORMSDIST(-0.14)" =2*0.4443= 0.8886
Answer:  The P-value is: 0.8886

Decision rule:
1) If p-value < level of significance (α) then we reject null hypothesis
2) If p-value > level of significance (α) then we fail to reject null hypothesis.
Here p value > 0.05 so we used second rule.
?Decision: Fail to Reject the null hypothesis.
 

Conclusion: There is not sufficient evidence to conclude that the percentage of offspring yellow peas is not? 25%.
Yes,the results contradict? expectations.