# Reducing the confidence interval with T distribition

My question is the following:

The population: weight of apples (in grams). We do not know anything about the population distribution except that it is a normal distribution.

We want to find a 90% confidence interval for the population mean.

So we take a random sample consists of 9 apples, their weights:

$$199, 198, 200,189, 186, 194, 179, 187, 196$$

Since it is a sample with small size $$n=9<30$$, and the standard deviation $$\sigma$$ is unknown, we need to estimate the variance, the standard deviation and the mean of the population and use the $$t$$ (student) distribution to calculate the confidence interval:

Estimator for the population mean is the sample mean:

$$\hat\mu=\overline X=\frac{1}{n}\sum_{i=1}^{n}x_i=192$$

Estimator for the population variance is the sample variance:

$$\hat\sigma^2=s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\overline X)^2=51$$

$$\hat\sigma=\sqrt {(s^2)}=s=7.141$$

We want 90% confidence interval:

$$1-\alpha=0.9 \Rightarrow \alpha=0.1$$

Calculation of the confidence interval using T distribution:

$$\overline X-\frac{s}{\sqrt{n}}t_{n-1, \frac{\alpha}{2}}\leqslant \mu \leqslant \overline X+\frac{s}{\sqrt{n}}t_{n-1, \frac{\alpha}{2}}$$

$$192-\frac{7.141}{\sqrt{9}}t_{8, \frac{0.1}{2}}\leqslant \mu \leqslant 192+\frac{7.141}{\sqrt{9}}t_{8, \frac{0.1}{2}}$$

$$t_{8, \frac{0.1}{2}}=1.860 \Rightarrow 187.5725 \leqslant\mu\leqslant 196.4274$$

Now we can say: We are 90% sure that the population mean ($$\mu$$) is in the range of: $$[187.5725, 196.4274].$$

We can see that the confidence interval has length of $$8.8549$$.

What is the problem? Confidence interval of $$8.8549$$ is too large for us, we want to reduce the size of the confidence interval to be at most of size $$4$$, that is $$2$$ for each side around the mean, and still keep the confidence level: 90%.

In other words: What will be the sample size if we want to reduce the size of the confidence interval to be at most $$4$$, with 90% confidence?

In mathematical terms, We should use this formula (to get the new sample size):

$$\frac{s}{\sqrt{n}}t_{n-1, \frac{0.1}{2}}\leqslant 2$$

My question is: how can i play with this equation to find the new sample size? i mean: I need to know what $$t_{n-1,\frac{0.1}{2}}$$ is at the time that $$n$$ is unknown.

It is the first time i work with such problems with $$t$$ distribution, I solved a lot of problems like that with $$z$$ distribution (when the sample size $$\gt 30$$ or $$\sigma$$ is known), with $$z$$ distribution there is no problem to do that because the $$z$$ value is function of the confidence level we want $$(\alpha)$$ only.

Thanks for reading the question so far, any suggestions?

Thanks!!!