## Question 432:

Asked on September 30, 2008

Tags:
t-score ,
TDIST

1

## Answer:

No answer provided yet.

- The distribution of x is said to be normally distributed with mean 68 and SD 3.75
*N(68,3.75). *It looks like were talking about all Male college students and it is the mean of this population (mu) that we're interested in.
- Calculate the z-score for 70 inches (70-68)/3.75 = .533. Now look this value up using the z-score to percentile calculator and select 1-sided area since we want to know the area greater than this point. You should get 70.3, so the area above this is 29.7%.
- X-bar is the sample mean of the male heights.
- You gave no sample data, so I have to assume the mean obtained is equal to the population mean. The standard error of the mean (SEM) is equal to the standard deviation divided by the square root of the sample size. You also didn't provide a sample standard deviation, assuming it's identical to the population SD (3.75) the SEM = 3.75/SQRT(16) = 3.75/4 = .9375
- The t-score for this value would be the same as calculated above .533, except there are 16-1 degrees of freedom. You can use Excel and type =TDIST(0.533,15,1) and you should get 30% (slightly higher than the 29.7% for part 2 above since it takes into account the smaller sample size).
- Calculate the z-score (67-68)/3.75 = -0.266. Since t and z are symmetrical, I just entered .266 =TDIST(0.533,15,1) b/c excel won't take a negative t-score. I got an area of 39.6% below 67.