Question

A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce.
Construct a 90% confidence interval for the population mean weight of the candies.

(i) State the confidence interval.
(ii) Sketch the graph.
(iii) Calculate the error bound.

Answer 1

Answer:

i) $2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959$  

$2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041$  

So then the 90% confidence interval is given by (1.959, 2.041)

ii) Figure attached

iii) $ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}$   (1)

And for a 90% of confidence the significance is given by $\alpha=1-0.9=0.1$, and $\frac{\alpha}{2}=0.05$. Since we know the population standard deviation we can calculate the critical value $z_{0.05}= \pm 1.64$

We know the folllowing data:

$\bar X = 2$ represent the sample mean

$\sigma = 0.1$ represent the population deviation

n =16 represent the sample size

Part i)

If we replace the values given into formula (1) we got:

$2 - 1.64 \frac{0.1}{\sqrt{16}}= 1.959$  

$2 + 1.64 \frac{0.1}{\sqrt{16}}= 2.041$  

So then the 90% confidence interval is given by (1.959, 2.041)

Part ii)

Figure attached. We have the illustration for the confidence interval obtained.

Part iii)

The margin of error is given by:

$ME=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}}$

And if we replace we got:

$ME= 1.64 *\frac{0.1}{\sqrt{16}}= 0.041$