Respuesta :
Answer:
Yes. There is 90% confidence the mean will be between ± 1.76% of the sample average.
Step-by-step explanation:
In this case, where the sample is little and the standard deviation of the population is unknown, the confidence interval (CI) can be expressed as
[tex][/tex][tex]\bar{x}\pm t_{n-1}*s/\sqrt{n}[/tex]
"s" is the standard deviation of the sample and t(n-1) is the critical t-value with (n-1) degrees of freedom. In this case, the degrees of freedom ar (4-1)=3.
The standard deviation is expressed as 1.5% or the mean, that is
[tex]s=0.015*\bar{x}[/tex]
Since we have two limits for the mean and a 90% CI, we have 5% of being outside the confidence. Looking up in the table for P(x>t)=0.05 and DF=3, we have
[tex]t_3=2.353[/tex]
Then the CI is
[tex]\bar{x} \pm 2.353*(0.015*\bar{x})/\sqrt{4}\\\\\bar{x} \pm 2.353*(0.015*\bar{x})/\sqrt{4}\\\\\bar{x} \pm 0.0176\bar{x}[/tex]
There is 90% confidence the mean will be between ± 1.76% of the sample average.
