Question

A survey of 1000 students found that 274 chose professional baseball team A as their favorite team. In a similar survey involving 760 students, 240 of them chose team A as their favorite. Compute a 95% con- fidence interval for the difference between the propor- tions of students favoring team A in the two surveys. Is there a significant difference?

Answer 1

Answer:

The 95% confidence interval would be given (-0.0851;0.00108).  

We are confident at 95% that the difference between the two proportions is between $-0.0851 \leq p_A -p_{A'} \leq 0.00108$

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 for brand A  

$\hat p_A =\frac{274}{1000}=0.274$ represent the estimated proportion for team A 1

$n_A=1000$ is the sample size

$p_{A'}$ represent the real population proportion for team A 2  

$\hat p_{A'} =\frac{240}{760}=0.316$ represent the estimated proportion for team A 2

$n_{A'}=760$ is the sample size required for team A 2

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

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  

$(\hat p_A -\hat p_{A'}) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_{A'} (1-\hat p_{A'})}{n_{A'}}}$  

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:  

$(0.274-0.316) - 1.96 \sqrt{\frac{0.274(1-0.274)}{1000} +\frac{0.316(1-0.316)}{760}}=-0.0851$  

$(0.274-0.316) + 1.96 \sqrt{\frac{0.274(1-0.274)}{1000} +\frac{0.316(1-0.316)}{760}}=0.00108$  

And the 95% confidence interval would be given (-0.0851;0.00108).  

We are confident at 95% that the difference between the two proportions is between $-0.0851 \leq p_A -p_{A'} \leq 0.00108$