Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 438 green peas and 173 yellow peas.

Answer 1

Answer:

a

The 95% confidence interval is $0.2474 < p < 0.3188$

b

  The result obtained does not contradict expectation

Step-by-step explanation:

From the question we are told that

   The number of green peas is  k =  438

   The number of yellow peas is u =  173

Generally the sample size is mathematically represented as

       $n = k + u$

=>    $n = 438 + 173$

=>    $n = 611$

Generally the sample proportion for yellow peas is

      $\^ p = \frac{u}{n}$

=>   $\^ p = \frac{173}{611 }$

=>   $\^ p = 0.2831$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

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

=>  $E = 1.96 * \sqrt{\frac{ 0.2831 (1- 0.2831)}{611} }$

=>  $E = 0.0357$

Generally 95% confidence interval is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>     $0.2831 -0.0357$

=>     $0.2474 < p < 0.3188$

From the question we are told that it was expected that 25% of the offspring peas will be yellow

   Now from the 95% confidence interval obtained we see that the expected sample proportion(25% ) falls within it so  it means that the result obtained does not contradict expectation