Question

The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the measurements of the sound intensity of a random sample of 9 such food processors showed a sample mean of 50.3 decibels, find a 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type.

Answer 1

Answer:

The 95% confidence interval would be given by (48.405;52.195)    

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=50.3$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=2.9$ represent the population standard deviation

n=9 represent the sample size  

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=2.9)$

We know from the central theorem that the distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

2) Confidence interval

The confidence interval for the mean is given by the following formula:

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$50.3-1.96\frac{2.9}{\sqrt{9}}=48.405$    

$50.3+1.96\frac{2.9}{\sqrt{9}}=52.195$    

So on this case the 95% confidence interval would be given by (48.405;52.195)