Question

Your friend has invented a random number generating machine that returns a value of 1, 2, 3, or 4 when a button is pressed, and you have been asked to test whether the machine displays each number 25% of the time. You generate 100 numbers and observe that 1, 2, and 3 appeared 24 times each and that 4 appeared 28 times. You decide to perform a test of mean of the random variable X, where X = 1 if 4 appears and X = 0 otherwise, and thus you have coded X = 1 for 28 of the observations and X = 0 for the other 72 observations. The null hypothesis is HO: the mean of X = 0.25, and the alternative hypothesis is H1: the mean of x > 0.25. You are preforming a one-tailed test. If X had been recorded as equal to 1 in only 25 observations (instead of 28), the standard deviation of X would have been 0.433013. What is the best excel formula to calculate the p-value for this test? =1-Normdist(0.28,0.25,0.0433013,True) O =1-Normdist(0.25,0.28,0.0433013,True) a =Normdist(0.25,0.28,0.433013,True) =1-Normdist(0.25,0.28,0.433013,True) O =1-Normdist(0.28,0.25,0.433013,True) O =Normdist(0.28,0.25,0.433013,True) =Normdist(0.28,0.25,0.0433013, True) =Normdist(0.25,0.28,0.0433013,True) O

Answer 1

Excel offers you the NORM.DIST statistical function for working with normal distributions. The NORM.DIST function calculates the probability that variable X falls below or at a specified value.

So, to find the probability that variable X falls above or at a specified value, we will have to use 1- normdist()

The NORM.DIST function uses the syntax:

=NORM.DIST(x,mean,standard_dev,cumulative)

Therefore, the answer is 1 - normdist( 0.28,0.25,0.433013,True)