Question

A recent study states that moonlight can affect the survival rate of new plants. It reported that 26 of 50 plants that were shielded from moonlight survived and 19 of 65 plants left unshielded survived. Choose the 95% confidence interval that represents the difference between the proportions of shielded and unshielded new plants that survive.

Answer 1

Answer:

$(0.52-0.292) - 1.96\sqrt{\frac{0.52(1-0.52)}{50} +\frac{0.292(1-0.292)}{65}}=0.0508$  

$(0.52-0.292) + 1.96\sqrt{\frac{0.52(1-0.52)}{50} +\frac{0.292(1-0.292)}{65}}=0.405$  

And the 95% confidence interval would be given (0.0508;0.405).  

We are confident at 95% that the difference between the two proportions is between $0.0508 \leq p_A -p_B \leq 0.405$

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".  

$p_A$ represent the real population proportion of shielded survived

$\hat p_A =\frac{26}{50}=0.52$ represent the estimated proportion of shielded survived

$n_A=50$ is the sample size required of shielded survived

$p_B$ represent the real population proportion for unshielded survived

$\hat p_B =\frac{19}{65}=0.292$ represent the estimated proportion for unshielded survived

$n_B=65$ is the sample size required for unshielded survived

$z$ represent the critical value for the margin of error  

Solution to the problem

The population proportion have the following distribution  

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

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.52-0.292) - 1.96\sqrt{\frac{0.52(1-0.52)}{50} +\frac{0.292(1-0.292)}{65}}=0.0508$  

$(0.52-0.292) + 1.96\sqrt{\frac{0.52(1-0.52)}{50} +\frac{0.292(1-0.292)}{65}}=0.405$  

And the 95% confidence interval would be given (0.0508;0.405).  

We are confident at 95% that the difference between the two proportions is between $0.0508 \leq p_A -p_B \leq 0.405$