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Course: AP®︎/College Statistics > Unit 11
Lesson 4: Confidence intervals for the difference of two means- Conditions for inference for difference of means
- Conditions for inference on two means
- Constructing t interval for difference of means
- Calculating confidence interval for difference of means
- Two-sample t interval for the difference of means (calculator-active)
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Calculating confidence interval for difference of means
Calculating confidence interval for difference of means.
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- Why does the formula using the formual in this video and in this unit, give different confidence intervals than using a 2-SampTInt on the TI-84? Is it possible to answer the questions in this section using the TI-84 functions or do they need to be answered by hand using the formula?(6 votes)
- if there is a "conservative" version of "degree of freedom", then is there a "non conservative" version of "degree of freedom"? if yes, how does it work?(3 votes)
- Yes; there are conservative and non-conservative ways to calculate the degrees of freedom (for a difference of two sample test);
1. Conservative; df = lower of (n_1 - 1) & (n_2 -1).
2. Non-Conservative; (s_1^2/n_1 + s_2^2/n_2)/((s_1^4/(n_1^2*(n_1-1))) + (s_2^4/(n_2^2*(n_2-1)))
As you can see it's very complicated and is usually only calculated using computers. :D
Hope this helps,
- Convenient Colleague(5 votes)
Video transcript
- [Instructor] Kylie
suspected that when people exercise longer, their
body temperatures change. She randomly assigned
people to exercise for 30 or 60 minutes, then
measured their temperatures. The 18 people who exercised for 30 minutes had a mean temperature, so this is the sample mean
for that sample of 18 folks, of 38.3 degrees Celsius,
with a standard deviation, this is a sample standard
deviation for those 18 folks, of 0.27 degrees Celsius. The 24 people who exercised 60 minutes had a mean temperature
of 38.9 degrees Celsius, with a standard deviation, this is, once again, these
are both sample means and sample standard deviations,
of 0.29 degrees Celsius. Assume that the conditions
for inference have been met, and that Kylie will use the
conservative degrees of freedom from the smaller sample size. Which of the following is
a 90% confidence interval for the difference in
mean body temperature after exercising for
the two amounts of time? So pause this video, and see
if you can figure it out. All right, now let's work
through this together. So, in previous videos, we talked about the general form of our
confidence interval, our t interval which we're going to use, because we're dealing with means, and we're dealing with
the differences in means. And so our t interval is
going to have the form: our difference between our sample means, so it could be the sample
mean for the 60 minute group minus the sample mean
for the 30 minute group. Plus or minus our critical t value, times our estimate of
the sampling distribution of the difference of the sample means. And that is going to be, I think I have enough space here to do it, that is going to be the
sample standard deviation of the 60 minute group squared over the sample size
of the 60 minute group, plus the sample standard deviation of the 30 minute group squared divided by the sample size
of the 30 minute group. And so, we can actually figure
out all of these things. So this is going to be equal to the sample mean for the
60 minute group is 38.9, so it's 38.9 minus the sample
mean for the 30 minute group, which is 38.3. 38.3. Plus or minus our critical t value. Now how do we figure that out? Well, we can use our 90% confidence level that we care about, this
90% confidence interval, but if we're looking
up things on a t table, we also need to know
our degrees of freedom. And it says here that Kylie will use the conservative degrees of freedom. And that means that she will
look at each of those samples, so one has a sample size of 18,
one has a sample size of 24. Whichever is lower, she
will use one less than that as her degrees of freedom. 18 is clearly lower than 24,
so the degrees of freedom in this situation is
18, or are 18 minus one, so 17. And so using that and that, we can now look this up on a t table. So our confidence level, 90%, and then our degrees of freedom,
17, so that is that row. The 90% confidence level is this column, and so that gives us our
critical t value of 1.74. So going back here, this is
going to be plus or minus 1.74 times the square root, times the square root. What's our sample standard deviation for the 60 minute group? Well, they give it right over here, 0.29, and we're gonna have to square that, divided by the sample size
for the 60 minute group, so let's see, the 24 people
who exercised for 60 minutes, so divided by 24. Plus the sample standard
deviation for the 30 minute group, so that's 0.27, 0.27 squared, divided by the sample size
for the 30 minute group, divided by 18. And we're done, and we can
look down at the choices. Let's see, they all got
the first part the same, 'cause that's maybe the
most straightforward part. 38.9 minus 38.3. Plus or minus 1.74, so both
of these are looking good, we can rule out these two 'cause they have a different critical t value. Now let's see, we have
0.29 squared divided by 24 plus .027 squared divided by
18, this one is looking good. Over here, let's see, they mixed up... they put the 30 minute sample size with the sample standard
deviation of the 60 minute group, so that won't work. And so we like choice A.