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AP®︎/College Statistics
Course: AP®︎/College Statistics > Unit 11
Lesson 5: Testing for the difference of two population means- Hypotheses for a two-sample t test
- Example of hypotheses for paired and two-sample t tests
- Writing hypotheses to test the difference of means
- Two-sample t test for difference of means
- Test statistic in a two-sample t test
- P-value in a two-sample t test
- Conclusion for a two-sample t test using a P-value
- Conclusion for a two-sample t test using a confidence interval
- Making conclusions about the difference of means
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Conclusion for a two-sample t test using a P-value
Given results of a two-sample t test, compare the P-value to the significance level to make a conclusion in context about the difference between two means.
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- Can someone explain to me why the DF is 197? I thought it was 99 variables + 99 variables would be 198?(11 votes)
- I think its not conservative df its that one which is calculated by complicated formula(1 vote)
- How come p-Value is 0.13
According to the null hypothesis, it is a two-tailed test. As it is a two-tailed test, the area of rejection falls both sides of the distribution. Shouldn't it be multiplied by 2 to get the actual p-value which is 0.26?(8 votes) - It shoudl be P value? caz italics_p suggests something else like proportions?(0 votes)
- please avoid often using long sentences like "..standard deviatio of sampling distribution of the sample mean" need not reiterate every time its one sample set from hypothetical sampling distribution(0 votes)
- I like such long sentences as it updates/refreshes my knowledge what it really is. It's always good the have a deep understanding and feeling what is "the full name".(3 votes)
Video transcript
- [Instructor] A sociologist
studying fertility in France and Switzerland wanted to test if there was a difference in
the average number of babies women in each country have. The sociologist obtained
a random sample of women from each country. Here are the results of their test. So you can see a 100-person
sample from France, a 100-person sample from Switzerland. They actually don't have
to be the same sample size. We have our sample means, our
sample standard deviations. You have the standard error of the mean, which for each sample
would be our estimate of the standard deviation
of the sampling distribution of the sample mean. And here it says t test for the means of these different
populations being different. And just to make sure we
can make sense of this, let's just remind
ourselves what's going on. The null hypothesis is
that there's no difference in the mean number of babies
that women in France have versus the mean number of babies that women in Switzerland have. That would be our null hypothesis, the no news here hypothesis. And our alternative would
be that they are different. And that's what we have right over here. It's a t test to see if we have evidence that would suggest our
alternative hypothesis. And so what we do is we
assume the null hypothesis. From that you're able to
calculate a t-statistic, and then from that t-statistic
and the degrees of freedom, you are able to calculate a p-value. And if that p-value is below
your significance level, then you'd say hey this was
pretty unlikely scenario, let me reject the null hypothesis, which would suggest the alternative. But if your p-value is greater
than your significance level, then you would fail to
reject your null hypothesis, and so you would not
have sufficient evidence to conclude the alternative. So what's going on over here? You really just have to compare this value to this value. It says at the alpha is equal
to 0.05 level of significance, is there sufficient evidence to conclude that there is a difference in
the average number of babies women in each country have? Well we can see that our p-value, 0.13, is greater than our alpha value, 0.05. And so because of that, we fail to reject our null hypothesis. And to answer their question, no there is not sufficient evidence to conclude that there is a difference. There's not sufficient evidence to reject the null hypothesis and
suggest the alternative.