Complete Question
A genetic experiment with peas resulted in one sample of offspring that consisted of 432 green peas and 164 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?
Answer:
The 95% confidence interval is [tex]0.2392 < p < 0.3108[/tex]
No, the confidence interval includes 0.25, so the true percentage could easily equal 25%
Step-by-step explanation:
From the question we are told that
The total sample size is [tex]n = 432 + 164 =596[/tex]
The number of offspring that is yellow peas is [tex]y = 432[/tex]
The number of offspring that is green peas is [tex]g = 164[/tex]
The sample proportion for offspring that are yellow peas is mathematically evaluated as
[tex]\r p = \frac{ 164 }{596}[/tex]
[tex]\r p = 0.275[/tex]
Given the the confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha = (100 - 95)\%[/tex]
[tex]\alpha = 5\% = 0.0 5[/tex]
The critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically evaluated as
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1- \r p )}{n} }[/tex]
=> [tex]E = 1.96 * \sqrt{\frac{0.275 (1- 0.275 )}{596} }[/tex]
=> [tex]E = 0.0358[/tex]
The 95% confidence interval is mathematically represented as
[tex]\r p - E < p < \r p + E[/tex]
=> [tex]0.275 - 0.0358 < p < 0.275 + 0.0358[/tex]
=> [tex]0.2392 < p < 0.3108[/tex]
