What happened to the normal condition np ≥ 10 and n(1-p) ≥ 10

it is for sampling distribution of sample proportion

Shouldn't the standard error of sample be just sample standard deviation instead of (sample standard deviation)/(sqrt(n)) ??

When you take a sample of a population, the sd should be sd/sqrt(n). What stays the same is the mean. The mean is the same both for the population and the sample. I think you're confusing the two.

It's said: *n should be >= 30*, for calculating t-intervals (using t-statistics). But in another video (https://www.khanacademy.org/math/statistics-probability/significance-tests-one-sample/more-significance-testing-videos/v/z-statistics-vs-t-statistics) Sal is saying that t-statistics should be used instead of z-statistics only *IF n < 30*.

z-statistics will give a narrower confidence interval than t-statistics, but the larger 𝑛 is the less that difference will be, and for 𝑛 ≥ 30, the difference can be considered negligible.

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Statistics and probability

Reference: Conditions for inference on a mean

Google Classroom

When we want to carry out inference (build a confidence interval or do a significance test) on a mean, the accuracy of our methods depends on a few conditions. Before doing the actual computations of the interval or test, it's important to check whether or not these conditions have been met. Otherwise the calculations and conclusions that follow may not be correct.

The conditions we need for inference on a mean are:

Random: A random sample or randomized experiment should be used to obtain the data.
Normal: The sampling distribution of $\bar{x}$ ‍ (the sample mean) needs to be approximately normal. This is true if our parent population is normal or if our sample is reasonably large $(n \geq 30)$ ‍.
Independent: Individual observations need to be independent. If sampling without replacement, our sample size shouldn't be more than $10 %$ ‍ of the population.

Let's look at each of these conditions a little more in-depth.

The random condition

Random samples give us unbiased data from a population. When we don't use random selection, the resulting data usually has some form of bias, so using it to infer something about the population can be risky.

[Want an example?]

More specifically, sample means are unbiased estimators of their population mean. For example, suppose we have a bag of ping pong balls individually numbered from

0

30

, so the population mean of the bag is

15

. We could take random samples of balls from the bag and calculate the mean from each sample. Some samples would have a mean higher than

15

and some would be lower. But on average, the mean of each sample will equal

15

. We write this property as

μ_{\bar{x}} = μ

, which holds true as long as we are taking random samples.

This won't necessarily happen if we use a non-random sample. Biased samples can lead to inaccurate results, so they shouldn't be used to create confidence intervals or carry out significance tests.

The normal condition

The sampling distribution of

\bar{x}

(a sample mean) is approximately normal in a few different cases. The shape of the sampling distribution of

\bar{x}

mostly depends on the shape of the parent population and the sample size

n

Case 1: Parent population is normally distributed

If the parent population is normally distributed, then the sampling distribution of

\bar{x}

is approximately normal regardless of sample size. So if we know that the parent population is normally distributed, we pass this condition even if the sample size is small. In practice, however, we usually don't know if the parent population is normally distributed.

Case 2: Not normal or unknown parent population; sample size is large ( $n \geq 30$ ‍)

The sampling distribution of

\bar{x}

is approximately normal as long as the sample size is reasonably large. Because of the central limit theorem, when

n \geq 30

, we can treat the sampling distribution of

\bar{x}

as approximately normal regardless of the shape of the parent population.

There are a few rare cases where the parent population has such an unusual shape that the sampling distribution of the sample mean

\bar{x}

isn't quite normal for sample sizes near

30

. These cases are rare, so in practice, we are usually safe to assume approximately normality in the sampling distribution when

n \geq 30

Case 3: Not normal or unknown parent population; sample size is small ( $n < 30$ ‍)

As long as the parent population doesn't have outliers or strong skew, even smaller samples will produce a sampling distribution of

\bar{x}

that is approximately normal. In practice, we can't usually see the shape of the parent population, but we can try to infer shape based on the distribution of data in the sample. If the data in the sample shows skew or outliers, we should doubt that the parent is approximately normal, and so the sampling distribution of

\bar{x}

may not be normal either. But if the sample data are roughly symmetric and don't show outliers or strong skew, we can assume that the sampling distribution of

\bar{x}

will be approximately normal.

The big idea is that we need to graph our sample data when $n < 30$ ‍ and then make a decision about the normal condition based on the appearance of the sample data.

The independence condition

To use the formula for standard deviation of

\bar{x}

, we need individual observations to be independent. In an experiment, good design usually takes care of independence between subjects (control, different treatments, randomization).

In an observational study that involves sampling without replacement, individual observations aren't technically independent since removing each observation changes the population. However the

10 %

condition says that if we sample

10 %

or less of the population, we can treat individual observations as independent since removing each observation doesn't change the population all that much as we sample. For instance, if our sample size is

n = 30

, there should be at least

N = 300

members in the population for the sample to meet the independence condition.

Assuming independence between observations allows us to use this formula for standard deviation of

\bar{x}

when we're making confidence intervals or doing significance tests:

σ_{\bar{x}} = \frac{σ}{\sqrt{n}}

We usually don't know the population standard deviation

σ

, so we substitute the sample standard deviation

s_{x}

as an estimate for

σ

. When we do this, we call it the standard error of

\bar{x}

to distinguish it from the standard deviation.

So our formula for standard error of

\bar{x}

is:

σ_{\bar{x}} \approx \frac{s_{x}}{\sqrt{n}}

Summary

If all three of these conditions are met, then we can we feel good about using

t

distributions to make a confidence interval or do a significance test. Satisfying these conditions makes our calculations accurate and conclusions reliable.

The random condition is perhaps the most important. If we break the random condition, there is probably bias in the data. The only reliable way to correct for a biased sample is to recollect the data in an unbiased way.

The other two conditions are important, but if we don't meet the normal or independence conditions, we may not need to start over. For example, there is a way to correct for the lack of independence when we sample more than

10 %

of a population, but it's beyond the scope of what we're learning right now.

The main idea is that it's important to verify certain conditions are met before we make these confidence intervals or do these significance tests.

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Brian Bale
Posted 5 years ago. Direct link to Brian Bale's post “What happened to the norm...”
What happened to the normal condition np ≥ 10 and n(1-p) ≥ 10
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(16 votes)
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- ronaldoamulya
  Posted 5 years ago. Direct link to ronaldoamulya's post “it is for sampling distri...”
  it is for sampling distribution of sample proportion
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  (42 votes)
Pramoth Viswan
Posted 6 years ago. Direct link to Pramoth Viswan's post “Shouldn't the standard er...”
Shouldn't the standard error of sample be just sample standard deviation instead of (sample standard deviation)/(sqrt(n)) ??
Button navigates to signup pageComment on Pramoth Viswan's post “Shouldn't the standard er...”
(8 votes)
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- Abe
  Posted 6 years ago. Direct link to Abe's post “When you take a sample of...”
  When you take a sample of a population, the sd should be sd/sqrt(n). What stays the same is the mean. The mean is the same both for the population and the sample. I think you're confusing the two.
  Button navigates to signup page
  (3 votes)
Alba Soma
Posted 2 years ago. Direct link to Alba Soma's post “It's said: *n should be >...”
It's said: n should be >= 30, for calculating t-intervals (using t-statistics). But in another video (https://www.khanacademy.org/math/statistics-probability/significance-tests-one-sample/more-significance-testing-videos/v/z-statistics-vs-t-statistics) Sal is saying that t-statistics should be used instead of z-statistics only IF n < 30.
Button navigates to signup pageComment on Alba Soma's post “It's said: *n should be >...”
(4 votes)
Answer
- Jerry Nilsson
  Posted 2 years ago. Direct link to Jerry Nilsson's post “z-statistics will give a ...”
  z-statistics will give a narrower confidence interval than t-statistics, but the larger 𝑛 is the less that difference will be, and for 𝑛 ≥ 30, the difference can be considered negligible.
  Comment on Jerry Nilsson's post “z-statistics will give a ...”
  (6 votes)
John Ostrowski
Posted 5 years ago. Direct link to John Ostrowski's post “I'm very curious about th...”
I'm very curious about the method for correcting the independence factor of samples when n> 10%N. It was mentioned as "beyond the scope", does anyone have references for that? Maybe Stat trek?
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Kyle Wright
Posted 3 years ago. Direct link to Kyle Wright's post “If the population is know...”
If the population is known to likely not be normal and a sample of n<30, then does the sample have to be transformed to normal to make an inference on CI? E.g. if you want to determine a CI on the number of customers that arrive at a drive through window during lunch - I believe this would be a Poisson counting process, therefore not normal. So wouldn't this be skewed to the right since the number of customers is bounded to >=0 and likely not symmetric. Basically, how can you correct a sample to make an inference on CI mean if dealing with Case #3. Thanks!
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(1 vote)
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