Answer:
we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )
Step-by-step explanation:
Given data
n=29
mean of x = 49.98 mm
S = 0.14 mm
μ = 50.00 mm
Cl = 95%
to find out
Can we be 95% confident that machine calibrated properly
solution
we know from t table
t at 95% and n -1 = 29-1 = 28 is 2.048
so now
Now for 95% CI for mean is
(x - 2.048 × S/√n , x + 2.048 × S/√n )
(49.98 - 2.048 × 0.14/√29 , 49.98 + 2.048 × 0.14/√29 )
( 49.926757 , 50.033243 )
hence we say for μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757 , 50.033243 )
The answer to your question is 30+4.+10+4
Answer:
Following are the solution to the given question:
Step-by-step explanation:
Please find the complete question in the attached file.
Arrange the value into the ascending order:
Calculating the Range:
Outlier = This might be the value "far" from other data values= 62
The range of scores would be, sans outlier,
Calculating the range without outlier
Inter Quartile Range
Calculating the IQR without outlier
Therefore,
Option a is "True".
Option b is "True".
Option c is "True".
Option d is "False".
Option e is "False".
Option f is "True".
Answer:
Expression: 2.89+(1.28x3) Answer: 6.73
Step-by-step explanation:
Number 1
The answer to this is 0.2725