The standard deviation of the sampling distribution of the sample proportion is equal to:

$$\frac{P\left(1-P\right)}{n}$$  

When computing the confidence interval for the population proportion, the Student's t-distribution is used rather than the normal distribution.

As the confidence level for a confidence interval increases, the width of the interval also increases.

Assuming unknown but equal population variances, determine the number of degrees of freedom for the following

$$n_1=30;\ s_1^2=16.5;\ n_2=45;\ s_2^2=17.2$$  

A $$100\left(1-\alpha\right)\%$$  confidence interval for the difference between two means, independent samples, and known population variances is given by:

$$\overline{x}-\overline{y}\pm z_{\frac{\alpha}{2}}\sqrt[]{\frac{\sigma_x^2}{n_x}-\frac{\sigma^2y}{n_y}}$$  

Interval estimates for the variance of a normal population rely on the random variable: $$\frac{\left(n-1\right)s^2}{\sigma^2}$$  , which follows........