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## Statistics and probability

### Course: Statistics and probability>Unit 11

Lesson 1: Introduction to confidence intervals

# Interpreting confidence level example

Interpreting confidence level example.

## Want to join the conversation?

• I understand that answer C is the best answer. But I don't understand why answer B is a false one.
If this method produces intervals that capture the true parameter in roughly 90% of the intervals cronstructed, why can not I say that any of these intervals I pick, including (341,359), has a 90% chance of containing the true parameter?

In the previous video "Confidence intervals and margin of error" at , if I got it right, it was said that these two statements are equivalent:
(i) The probability that p̂=0.54 is within 2*sd(p̂) is 95%.
(ii) There is a 95% probability that p is within 2*sd(p̂) of p̂=0.54.
In that case, (0.54-2*sd(p̂) , 0.54+2*sd(p̂)) is one of the many intervals the repeatedly sampling could produce.
Isn't (ii) analogous to say that there is 90% probabilty that the true mean is within (341,359)? •  Think of it this way: if you have a single sample with interval that doesn't overlap with the true population parameter then would it be reasonable to assume this particular interval contains the true parameter 90% of the times? Obviously no, because no matter how many times you repeat the experiment with this particular interval, the true parameter won't be captured. So that's why you take multiple samples with their respective intervals and in 90% of all the samples, those different intervals contain the true parameter.
• Example 2 in the next lesson "Interpreting confidence levels and confidence intervals" addresses the B choice. It seems that the problem is with wording. As far as I understand according to it we should say "We're 90% confident the interval captured the true mean" as opposed to "There's a 90% chance the interval captured the true mean". In my mind the two wordings are equivalent though.

If choice C is a better interpretation and choice B is not necessarily false I think it'd be better if the exercise asked to choose the best interpretation, not the "correct" one.

"Can we say there is a 95% chance that the true mean is between 110 and 120 kilometers per hour?

We shouldn't say there is a 95% chance that this specific interval contains the true mean, because it implies that the mean may be within this interval, or it may be somewhere else. This phrasing makes it seem as if the population mean is variable, but it's not. This interval either captured the mean or didn't. Intervals change from sample to sample, but the population parameter we're trying to capture does not.
It's safer to say we're 95%, percent confident that this interval captured the mean, since this phrasing more closely agrees with the long-term capture rate of confidence levels." • I think you shouldn't use the words "chance/probability" instead of "confident", Because it implies that the true mean is variable.
Consider this:
There is a 16.67% chance of getting a 1 when roll a die.
This means whenever you roll a die, the number on top of the die will change or vary each time (unlike the true mean, which is fixed).

Like you said, the problem is with the wording. It is one of my weaknesses in mathematics :(
• 90% confidence interval is within 1.65 standard deviation, if 90% is being between (341, 359) then how can standard deviation be 25? (1.65 * 25 ≈ 41.25) • All the answers seem right to me :/ • Wish Sal makes a video on why choice B is false • According to choice B, You use the words "probability/chance" to an event that has not been determined yet (such as a probability of rolling a die and get 4). But both true mean and interval already have been determined, So it is not about probability anymore.

Am I correct? • why is choice B wrong? • Don't we need to know the population SD to create a confidence interval for sampling dists of sampling means?

Here we have a sample where
n = 30
sample mean = 350
sample SD = 25

And the sampling dist of sample means where n = 30 is approx normal since n is big enough, and has a mean of the true population mean

Using a z-table, we know that there is a 90% chance that a sample mean will land within 1.64 SD of sample means from the true pop mean.
So our 90% confidence interval will look like this:
(350 - 1.64*SD, 350 + 1.64*SD)

SD of sample means is SD of population divided by n (30). But we don't have SD of population, only a sample SD of 25, so we can use the sample SD instead. But the sample SD is a biased estimator of SD pop, so won't out confidence interval be biased as well?   