Respuesta :
Answer:
a) The sample mean is of 49 and the sample standard deviation is of 11.7.
b) The range of the true mean at 90% confidence level is of 9.62 hours.
c) The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.
Step-by-step explanation:
Question a:
Sample mean:
[tex]\overline{x} = \frac{34+40+46+49+61+64}{6} = 49[/tex]
Sample standard deviation:
[tex]s = sqrt{\frac{(34-49)^2+(40-49)^2+(46-49)^2+(49-49)^2+(61-49)^2+(64-49)^2}{5}} = 11.7[/tex]
The sample mean is of 49 and the sample standard deviation is of 11.7.
b)Determine the range of the true mean at 90% confidence level.
We have to find the margin of error of the confidence interval. Since we have the standard deviation for the sample, the t-distribution is used.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 6 - 1 = 5
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 2.0.150
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}}[/tex]
In which s is the standard deviation of the sample and n is the size of the sample. So
[tex]M = 2.0150\frac{11.7}{\sqrt{6}} = 9.62[/tex]
The range of the true mean at 90% confidence level is of 9.62 hours.
(c)If a seventh sample is tested, what is the prediction interval (90% confidence level) of its failure time.
This is the confidence interval, so:
The lower end of the interval is the sample mean subtracted by M. So it is 49 - 9.62 = 39.38 hours.
The upper end of the interval is the sample mean added to M. So it is 49 + 9.62 = 58.62 hours.
The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.
