The null and alternate hypotheses are: H0: 1 2 H1: 1 2 A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard devi- ation of 15. At the .10 significance level, is there a difference in the population means?

The degreess of freedom are given by:

Now we can calculate the p value with the following probability:

For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.

Step-by-step explanation:

Information provided

represent the mean for sample 1  

represent the mean for sample 2  

represent the sample standard deviation for 1  

represent the sample standard deviation for 2  

sample size for the group 2  

sample size for the group 2  

Significance level provided

t would represent the statistic

Hypothesis to test

We want to verify if the true means for this case are significantly different, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

The statistic is given by:

(1)  

And the degrees of freedom are given by  

Replacing the info given we got:

The degreess of freedom are given by:

Now we can calculate the p value with the following probability:

For this case since the p value is higher than the significance level we can FAIL to reject the null hypothesis and we can conclude that the true means are NOT significantly different at 10% of significance.

um, this was months ago lol!

Step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.