It is hypothesized that 50% of the U.S. population cannot name both of the U.S. Senators from the state in which they reside. A sample of 500 individuals is taken and 218 correctly name both of their Senators. What decision is made at the 1% level of significance and what test statistic was used?

Hello Anne,

First of all, when it comes to statistic, the letters I am using may be different from what your class is using, but the formulas will be the same.

Sine we are dealing with a comparison of proportions between a population and sample, , we need to do a z-test using the formula:

z=(p-π)/σ

π(pi) = population proportion or probability (.5)

p = sample proportion of probability (218/500=.436)

σ = standard deviation

The formula to find standard deviation is:

σ = √(π(π-1)/n)

= √(.5(1-.5)/500

= √(.25/500)

= .02236

We take this back to our z formula.

z = (p-π)/σ

= (.436-.5)/.02236

= (-.064/.02236)

= -2.862

The level of confidence we are given is α=.01. We will use a two tailed test since no direction of difference was expected, meaning .005 in either tail. By looking at your z distribution, you can see that this corresponds to a z score of ±2.58.

Our calculated z of -2.862 is more extreme than the critical z of ±2.58, we reject the null hypothesis. The conclusion is that the proportion is a value other than 50%.

Anthony