2122 - Midterm 2 Review

The variance of the sampling distribution of sample means is equal to the variance of the population mean divided by the square root of the sample size.

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.

Suppose that: n=675  and sample proportion = 0.1. Find the UCL of the confidence interval for estimating the population proportion for a 90% confidence interval.

(Round to the nearest 2 decimal points)

(a)  

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)  

$$\overline{d}=19.56;\ s_d=2.5;\ n=20$$  

Calculate the margin of error for a 95% confidence interval using the above data.

(Round your answer to 2 decimal points)

(a)  

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

According to the central limit theorem, the sampling distribution of sample means can be approximated by the normal distribution as the number of samples gets large enough.

If the finite population correct factor is 0.6869 and the population size is 100, then the sample size is

(a)