Question

genetic experiment with peas resulted in one sample of offspring that consisted of green peas and yellow peas. a. Construct a ​% 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 ​% confidence interval. Express the percentages in decimal form. nothingp nothing ​(Round to three decimal places as​ needed.) b. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations? ​No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25% ​Yes, the confidence interval does not include​ 0.25, so the true percentage could not equal​ 25%

Answer 1

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  $0.2392 < p < 0.3108$

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  $n = 432 + 164 =596$

   The  number of  offspring that is yellow peas is $y = 432$

   The  number of  offspring that is green peas   is $g = 164$

   

The sample proportion for offspring that are yellow peas is mathematically evaluated as

        $\r p = \frac{ 164 }{596}$

        $\r p = 0.275$

Given the the  confidence level is  95% then the level of significance is mathematically represented as

       $\alpha = (100 - 95)\%$

      $\alpha = 5\% = 0.0 5$

The  critical value of  $\frac{\alpha }{2}$ from the normal distribution table is  

      $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically evaluated as

        $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1- \r p )}{n} }$

=>      $E = 1.96 * \sqrt{\frac{0.275 (1- 0.275 )}{596} }$

=>      $E = 0.0358$

The  95%  confidence interval is mathematically represented as

      $\r p - E < p < \r p + E$

=>   $0.275 - 0.0358 < p < 0.275 + 0.0358$

=>   $0.2392 < p < 0.3108$