Project 38: One Prop Z-Test

Suppose a carnival director claims that contestants have a 60% chance of winning a certain game at the carnival.  We believe that the true proportion of winning the game is less than this and plan to test our hypothesis at the .05 level.

Ho: p = .6

Ha: p < .6

We decide to play the game 60 times and we win 28 times.

phat = 28/60 = .467

Explanation: If p really were .6 and we were to play 60 times, the sampling distribution of phat would be approximately normal (60*.6 = 36 >10 AND 60*.4 = 24 >10).  The mean of the sampling distribution would be .6 and the standard deviation would be sqrt(.6*.4/60) = .063

The p-value for the test would be .0174.  This is less than .05 so we would reject Ho and get evidence that the true proportion of times a contestant would win this game is less than .6

Project 38: Variables 'po', 'x', 'n', 'haDirection' have been initialized.   Working methods called normalcdf and invNorm are available.

po represents the hypothesized proportion

n represents the sample size

x represents the number of successes in the sample

haDirection represents the direction of the alternative hypothesis (1:<, 2: >, 3 ≠) 

Review projects 2-5 if you need a reminder of how normalcdf and invNorm work.

Task: Initialize values for 'z' and 'pValue' that represent the test statistic (z-score) and p-value for the given test.

**If your code works for 5 test cases, you can enter your e-mail address

Universal Computational Math Methods:

pow(5,2) returns 25.0

abs(-3.0) returns 3

sqrt(49.0) returns 7.0