– [Instructor] Miriam was

testing her null hypothesis that the population mean of some data set is equal to 18 versus her

alternative hypothesis is that the mean is less than 18 with a sample of seven observations. Her test statistic, I

can never say that right, was t is equal to negative 1.9. Assume that the conditions

for inference were met. What is the approximate p

value for Miriam’s test? So, pause this video and see if you can figure this out on your own. Alright, what I always like

to remind ourselves what’s going on here before I go ahead

and calculate the p value. There’s some data set,

some population here and the null hypothesis is

that the true mean is 18, the alternative is that it’s less than 18. To test that null hypothesis,

Miriam takes a sample, sample size is equal to seven. From that, she would

calculate her sample mean and her sample standard deviation, and from that, she would

calculate this t statistic. The way she would do that or if they didn’t tell us ahead

of time what that was. We would say the t statistic

is equal to her sample mean, minus the assumed mean

from the null hypothesis, that’s what we have over here, divided by and this is a mouthful, our approximation of the standard error of the mean. The way we get that approximation, we take our sample standard deviation and divide it by the square

root of our sample size. Well, they’ve calculated

this ahead of time for us. This is equal to negative 1.9. So, if we think about a t distribution, I’ll try to hand draw a rough

t distribution really fast, and if this is the mean

of the t distribution, what we are curious about,

because our alternative hypothesis is that the

mean is less than 18. What we care about is,

what is the probability of getting a t value that

is more than 1.9 below the mean so this right

over here, negative 1.9. It’s this area, right there. I’m gonna do this with a TI-84, at least an emulator of a TI-84. All we have to do is, we

would go to 2nd distribution and then I would use the t cumulative distribution function so let’s go there, that’s the number six

right there, click enter. My lower bound… Yeah, I essentially wanted

it to be negative infinity and we can just call

that negative infinity. It’s an approximation of negative infinity, very, very low number. Our upper bound would be

negative 1.9, negative 1.9. And then our degrees of freedom, that’s our sample size minus one. Our sample size is seven so our degrees of freedom would be six. There we have it. This would be, our p value

would be approximately 0.053. Our p value would be approximately 0.053. Then what Miriam would do is,

would compare this p value to her preset significance

level, to alpha. If this is below alpha,

then she would reject her null hypothesis, which

would suggest the alternative. If this is above alpha, then she would fail to reject her null hypothesis.

Thanks, even thought I’m in 8th grade and have so little understanding on this

I tried to follow the same steps as you. I'm using a TI-83 Plus calculator. I don't understand why I got different results. I did tcdf(-1^99,-1.9, 6) and the answer came out to about -.12487…. I tried tcdf(-10^99, -1.9, 6) and then it gave me the same answer you guys got. Is there some function/setting on my calculator that I need to change?