# Normal Distribution Or Student Distribution?

If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value. For statistical purposes, you decide to throw $$10$$ fifty-cent coins in vending machine A. This results in a sample mean of $$7.49$$ $$g$$ and a sample variance of $$0.011$$ $$g^2$$. You may assume a normal distribution as model distribution for the measurements.
Construct a $$95%$$ confidence interval for the mean weight $$μ_A$$ of a fity-cent coin thrown in machine $$A$$.

In this exercise, you'll use the student distribution or the standard normal distribution to construct the confidence interval? I know that they say 'You may assume a normal distribution as model distribution for the measurements' but they also give you the sample variance (that is used with the student distribution) and not the variance. Furthermore, $$n=10$$ is quite small, so, in my opinion, I should use the student distribution, it's correct?

• Is the question which distribution to use part of the exercise? – Klaas van Aarsen Jan 22 at 22:17
• No but in order to find the confident interval I have to choose one, so I know if I have to use the z or t value. – Mark Jacon Jan 22 at 22:35
• The remark about the normal distribution could mean that the underlying distribution is normal, which is a requirement for the sample distribution to be a t-distribution. This is important for small sample sizes. So yes, you should base the confidence interval on the t-distribution. – Klaas van Aarsen Jan 23 at 7:31

You are correct that you need to use Student's t distribution with $$n - 1$$ $$= 10 - 1 = 9$$ degrees of freedom. That's because the population variance $$\sigma^2$$ is estimated by the sample variance $$S^2 = 0.011.$$ (Strictly speaking, the sample size is not relevant to the decision to use the t distribution instead of the normal distribution.)

The formula to get a 95% confidence interval for the population mean $$\mu$$ is $$\bar X \pm t^*\sqrt{S^2/n},$$ where $$t^*$$ cuts area 2.5% from the upper tail of $$\mathsf{T}(\text{df}=9).$$ Using, R statistical software, I get the value for $$t^*$$ shown below. If your course uses printed tables, you should look at the appropriate table now to find essentially the same number (typically, with fewer decimal places).

qt(.975, 9)
## 2.262157


Notes: For a 95% CI and a sample of size $$n = 35,$$ software gives $$t^* =2.0322,$$ as shown below. If you were given the exact value of the population variance $$\sigma^2,$$ then you could use the formula $$\bar X \pm 1.96\sqrt{\sigma^2/n},$$ where the value $$z^* = 1.960$$ cuts 2.5% of the area from the upper tail of a standard normal distribution.

qt(.975, 34)
## 2.032245
qnorm(.975)
## 1.959964


For $$n > 30$$ and 95% confidence intervals, both $$t^*$$ and $$z^* =1.960$$ round to $$2.0.$$ That's why some of the less-fussy textbooks say you can use $$z^*$$ instead of $$t^*$$ for $$n > 30.$$ However, this "rule of 30" works only for 95% confidence intervals.

• thanks a lot, now it's clear! – Mark Jacon Jan 23 at 17:19