The Sampling Distribution of the Sample Mean

Statistics
Tutor: None Selected Time limit: 1 Day

Mar 10th, 2015

We have a normal distribution with mean = 4.2 and sd =1.3.

a. So the probability x>5 is given by:

1-norm.cdf(5, mean = 4.2, sd = 1.3) = 1-0.73085 = 0.26915

or equivalently, p(z>(5-4.2)/1.3).


b. We have a sample mean = 5, sample size n = 8. Since 8 is very small we use the t-distribution to estimate sample mean distributions, with df = n-1 = 7. 

The standard error of our sample mean is sd/sqrt(n)

= 1.3/sqrt(7)

0.491353815

So our t-stat is given by (sample mean - population mean) / standard error

t = (5-4.2) / 0.49

= 1.628


Therefore, the probability our sample mean is greater than 5 will be one minus the probability it is less, which is

1 - P(t = 1.628, df = 7) 

= 1 - 0.926241741

= 0.0738

Mar 10th, 2015

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