Respuesta :
Answer:
The 90% confidence interval to estimate the mean breaking weight for this type cable is between 751.38 lb and 803.42 lb.
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the student t-distribution to solve this question.
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 = 47 - 1 = 46
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 46 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.6787
The margin of error is:
M = T*s = 1.6787*15.5 = 26.02 lb
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 777.4 - 26.02 = 751.38
The upper end of the interval is the sample mean added to M. So it is 777.4 + 26.02 = 803.42
The 90% confidence interval to estimate the mean breaking weight for this type cable is between 751.38 lb and 803.42 lb.
Answer:
[tex]777.4-1.679\frac{15.5}{\sqrt{47}}=773.60[/tex]
[tex]777.4+1.679\frac{15.5}{\sqrt{47}}=781.20[/tex]
We are 90% confident that the true mean for the breaking weigth for this typpe of cable is between 773.60 and 781.20
Step-by-step explanation:
Data provided from the problem
[tex]\bar X=777.4[/tex] represent the sample mean for the breaking weights
[tex]\mu[/tex] population mean
s=15.5 lb represent the sample standard deviation
n=47 represent the sample size
Confidence interval
The confidence interval for the true population mean is given by:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom, given by:
[tex]df=n-1=47-1=46[/tex]
The Confidence level is 0.90 or 90%, the significance would be [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex] and the critical value for this case is [tex]t_{\alpha/2}=1.679[/tex]
Replacing into the confidence formula we got:
[tex]777.4-1.679\frac{15.5}{\sqrt{47}}=773.60[/tex]
[tex]777.4+1.679\frac{15.5}{\sqrt{47}}=781.20[/tex]
We are 90% confident that the true mean for the breaking weigth for this typpe of cable is between 773.60 and 781.20