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The z-test requires the population standard deviation.

$$ z=\dfrac{ \bar x -\mu_0 }{\sigma/\sqrt n} $$

You don’t estimate $\sigma$ from the data; you know it. If this sounds unreasonable, you’re right,$^{\dagger}$ so we have the t-test, which uses the sample standard deviation, which is calculated from the data.

$$ t=\dfrac{ \bar x-\mu_0 }{s/\sqrt n}\\ s=\dfrac{1}{n-1}\sum\bigg( x_i-\bar x \bigg)^2 $$

$^{\dagger}$There are situations where this is reasonable, but I would consider them the exception.

We can simulate this to show that the equations give different values.

set.seed(2022);
n <- 31;
true_mean <- 0.2;
mu_0 <- 0;
true_sd <- 1;
x <-rnorm(n, true_mean, true_sd);
z_stat <- (mean(x) - mu_0)/(true_sd/sqrt(n));
t_stat <- (mean(x) - mu_0)/(sd(x)/sqrt(n));
z_stat - t_stat

I get a difference of about $0.05$.

REFERENCES

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.3

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.1

The z-test requires the population standard deviation.

$$ z=\dfrac{ \bar x -\mu_0 }{\sigma/\sqrt n} $$

You don’t estimate $\sigma$ from the data; you know it. If this sounds unreasonable, you’re right,$^{\dagger}$ so we have the t-test, which uses the sample standard deviation, which is calculated from the data.

$$ t=\dfrac{ \bar x-\mu_0 }{s/\sqrt n}\\ s=\dfrac{1}{n-1}\sum\bigg( x_i-\bar x \bigg)^2 $$

$^{\dagger}$There are situations where this is reasonable, but I would consider them the exception.

We can simulate this to show that the equations give different values.

set.seed(2022);
n <- 31;
true_mean <- 0.2;
mu_0 <- 0;
true_sd <- 1;
x <-rnorm(n, true_mean, true_sd);
z_stat <- (mean(x) - mu_0)/(true_sd/sqrt(n));
t_stat <- (mean(x) - mu_0)/(sd(x)/sqrt(n));
z_stat - t_stat

I get a difference of about $0.05$.

Addressing the question in the title, if you know variance or standard deviation, you know the other by either squaring (to get the variance from the standard deviation) or taking the square root (to get the standard deviation from th e variance).

REFERENCES

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.3

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.1

The z-test requires the population standard deviation.

$$ z=\dfrac{ \bar x -\mu_0 }{\sigma/\sqrt n} $$

You don’t estimate $\sigma$ from the data; you know it. If this sounds unreasonable, you’re right,$^{\dagger}$ so we have the t-test, which uses the sample standard deviation, which is calculated from the data.

$$ t=\dfrac{ \bar x-\mu_0 }{s/\sqrt n}\\ s=\dfrac{1}{n-1}\sum\bigg( x_i-\bar x \bigg)^2 $$

$^{\dagger}$There are situations where this is reasonable, such as when you are dealing with a proportion variable.

We can simulate this to show that the equations give different values.

set.seed(2022);
n <- 31;
true_mean <- 0.2;
mu_0 <- 0;
true_sd <- 1;
x <-rnorm(n, true_mean, true_sd);
z_stat <- (mean(x) - mu_0)/(true_sd/sqrt(n));
t_stat <- (mean(x) - mu_0)/(sd(x)/sqrt(n));
z_stat - t_stat

I get a difference of about $0.05$.

REFERENCES

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.3

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.1

The z-test requires the population standard deviation.

$$ z=\dfrac{ \bar x -\mu_0 }{\sigma/\sqrt n} $$

You don’t estimate $\sigma$ from the data; you know it. If this sounds unreasonable, you’re right,$^{\dagger}$ so we have the t-test, which uses the sample standard deviation, which is calculated from the data.

$$ t=\dfrac{ \bar x-\mu_0 }{s/\sqrt n}\\ s=\dfrac{1}{n-1}\sum\bigg( x_i-\bar x \bigg)^2 $$

$^{\dagger}$There are situations where this is reasonable, but I would consider them the exception.

We can simulate this to show that the equations give different values.

set.seed(2022);
n <- 31;
true_mean <- 0.2;
mu_0 <- 0;
true_sd <- 1;
x <-rnorm(n, true_mean, true_sd);
z_stat <- (mean(x) - mu_0)/(true_sd/sqrt(n));
t_stat <- (mean(x) - mu_0)/(sd(x)/sqrt(n));
z_stat - t_stat

I get a difference of about $0.05$.

REFERENCES

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.3

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.1

The z-test requires the population standard deviation.

$$ z=\dfrac{ \bar x -\mu_0 }{\sigma/\sqrt n} $$

You don’t estimate $\sigma$ from the data; you know it. If this sounds unreasonable, you’re right,$^{\dagger}$ so we have the t-test, which uses the sample standard deviation, which is calculated from the data.

$$ t=\dfrac{ \bar x-\mu_0 }{s/\sqrt n}\\ s=\dfrac{1}{n-1}\sum\bigg( x_i-\bar x \bigg)^2 $$

$^{\dagger}$There are situations where this is reasonable, such as when you are dealing with a proportion variable.

REFERENCES

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.3

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.1

The z-test requires the population standard deviation.

$$ z=\dfrac{ \bar x -\mu_0 }{\sigma/\sqrt n} $$

You don’t estimate $\sigma$ from the data; you know it. If this sounds unreasonable, you’re right,$^{\dagger}$ so we have the t-test, which uses the sample standard deviation, which is calculated from the data.

$$ t=\dfrac{ \bar x-\mu_0 }{s/\sqrt n}\\ s=\dfrac{1}{n-1}\sum\bigg( x_i-\bar x \bigg)^2 $$

$^{\dagger}$There are situations where this is reasonable, such as when you are dealing with a proportion variable.

We can simulate this to show that the equations give different values.

set.seed(2022);
n <- 31;
true_mean <- 0.2;
mu_0 <- 0;
true_sd <- 1;
x <-rnorm(n, true_mean, true_sd);
z_stat <- (mean(x) - mu_0)/(true_sd/sqrt(n));
t_stat <- (mean(x) - mu_0)/(sd(x)/sqrt(n));
z_stat - t_stat

I get a difference of about $0.05$.

REFERENCES

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.3

https://online.stat.psu.edu/stat200/lesson/8/8.2/8.2.3/8.2.3.1