Question

A low-noise transistor for use in computing products is being developed. It is claimed that the mean noise level will be below the 2.5-dB level of products currently in use. It is believed that the noise level is approximately normal with a standard deviation of .8. find 95% CI

Answer 1

The 95% CI is
`2.108 < \mu < 2.892`

Explanation:

From the question we are told that

The population mean
`\mu = 2.5`

The standard deviation is
`\sigma = 0.8`

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

`\alpha = 100 - 95`

=>
`\alpha = 5\%`

=>
`\alpha = 0.05`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution table, the values is
`Z_{(\alpha )/(2) } = 1.96`

Generally the margin of error is mathematically evaluated as

`E = Z_{(\alpha )/(2) } * (\sigma)/(√(n) )`

here we would assume that the sample size is n = 16 since the person that posted the question did not include the sample size

So

`E = 1.96* (0.8)/(√(16) )`

`E = 0.392`

The 95% CI is mathematically represented as

`\= x -E < \mu < \= x +E`

substituting values

`2.5 - 0.392 < \mu < 2.5 + 0.392`

substituting values

`2.108 < \mu < 2.892`