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?

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    $\begingroup$ Is the question which distribution to use part of the exercise? $\endgroup$ – Klaas van Aarsen Jan 22 at 22:17
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    $\begingroup$ 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. $\endgroup$ – Mark Jacon Jan 22 at 22:35
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    $\begingroup$ 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. $\endgroup$ – 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
## 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.

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

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