How Do You Calculate Sample Size for a T-Test with a 95% Confidence Interval?

[markosheehan]

View attachment 7546

i am stuck on question 12

I am trying to use the hypothesis testing 1 sample t test formula .

the right answer is 338 but I can not get this

 

[Jameson]

I think this question has to do with the lower bound of the confidence interval since the mean of the sample is above 100. What is the formula you've been given to calculate a confidence interval of a mean? What values do we already have that fit into that equation?

 

[markosheehan]

So the forumula is x-u/(s/✓n). Where X is the value 101.6

U is the mean which is 100,s is the standard deviation which is 15 , n is the size of the sample which is what we are looking for. What I'm not sure is what I am supposed to equal this to. I know it's something to do with the 5% level of confidence, but that forumula gives its answer as a z value.

 

[Jameson]

So the forumula is x-u/(s/✓n). Where X is the value 101.6

U is the mean which is 100,s is the standard deviation which is 15 , n is the size of the sample which is what we are looking for. What I'm not sure is what I am supposed to equal this to. I know it's something to do with the 5% level of confidence, but that forumula gives its answer as a z value.

On the right track!

So here's our confidence interval formula: $$\overline{x} \pm t^{*} \frac{s}{\sqrt{n}}$$.

Like you wrote, we just care about the lower bound so we can ignore the upper bound for now. The $t$ here corresponds to the 95% confidence level. Usually we say when $n>30$ we can use the normal distribution which would make $t=1.96$. If we plug that in we get:

$$100 = 101.6 -1.96 \frac{15}{\sqrt{n}}$$

Now we want to solve for $n$.