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:
`(48.41,\ 52.19)`

Step-by-step explanation:

The confidence interval for population mean is given by :-

`\overline{x}\pm z_(\alpha/2)(\sigma)/(√(n))`

Given : Sample size :
`n=9`

Sample mean :
`\ovreline{x}=50.3\text{ decibels}`

Standard deviation :
`\sigma=2.9\text{ decibels }`

Significance level :
`\alpha=1-0.95=0.05`

Critical value :
`z_(\alpha/2)=z_(0.025)=1.96`

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

`50.3\pm (1.96)(2.9)/(√(9))\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)`