Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 423 green peas and 159 yellow peas. a. Construct a 90​% 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 1

Answer:

a) $0.273 - 1.64\sqrt{\frac{0.273(1-0.273)}{582}}=0.243$

$0.273 + 1.64\sqrt{\frac{0.273(1-0.273)}{582}}=0.303$

The 90% confidence interval would be given by (0.243;0.303)

b) For this case since the confidence interval contains the value of 0.25 we don't have enough evidence to conclude that the true proportion of yellow peas is not 25%, so then the null hypothesis that the value expected of 25% for yellow peas can't be rejected at 10% of significance.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

Part a

We can estimate the proportion of yellow peas like this:

$\hat p = \frac{159}{423+159} = 0.273$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.273 - 1.64\sqrt{\frac{0.273(1-0.273)}{582}}=0.243$

$0.273 + 1.64\sqrt{\frac{0.273(1-0.273)}{582}}=0.303$

The 90% confidence interval would be given by (0.243;0.303)

Part b

For this case since the confidence interval contains the value of 0.25 we don't have enough evidence to conclude that the true proportion of yellow peas is not 25%, so then the null hypothesis that the value expected of 25% for yellow peas can't be rejected at 10% of significance.