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

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

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, (\sigma)/(√(n)))`

2) Confidence interval

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(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(2.9)/(√(9))=48.405`

`50.3+1.96(2.9)/(√(9))=52.195`

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