A random sample of 42 mechanic
A random sample of 42 mechanical gears has been collected, andthe measurements ofGear-Diameter (mm) are found below:130 131 132 130.5 130 130.2 130.7 130 131.2 133130 130 131 132 133 133 132 131 312.5 131.5131.2 132.5 130.2 129.5 131 130 130.2 131.5 131.4 131.6130.6 130.5 131.1 130 13 130 131 130 130 132131.2 132(a) Construct a 94% two-bounds confidence interval on thediameter.(b) Construct a 94% lower confidence bound on the diameter.
please solve manually
Answer:
A random sample of 42 mechanical gears is given below
130 131 132 130.5 130 130.2 130.7 130 131.2 133130 130 131 132 133 133 132 131 312.5 131.5131.2 132.5 130.2 129.5 131 130 130.2 131.5 131.4 131.6130.6 130.5 131.1 130 13 130 131 130 130 132131.2 132
[ Note that one observation is ” 13 ” which looks likeincorrect entry , but due to no evidence we will continuecalculation with considering “13” as correct entry ]
Here n = 42 ( sample size )
Now we will calculate sample mean 
and sample variance s2
= 
= ( 130 +131 +132 +130.5 +130 + …….. + 130 +132+131.2 +132 ) /42
= 5565.1 / 42
= 132.5024
And
s2 = 
= { (130-132.5024)2 +(131-132.5024)2+(132-132.5024)2 +……. + (131.2-132.5024)2+(132-132.5024)2 } / (42-1)
s2 = 46806.77 / 41
s2 = 1141.629
Thus sample standard deviation s = 
= 33.78799
So we have
= 132.5024 , s = 33.78799
(a) Construct a 94% two-bounds confidence interval on thediameter.
94% two-bounds confidence interval CI is given by
CI = { – 
* 
, – * }
where is t-distributed with n-1 = 42-1 = 41 degree of freedom (DF) and
= 0.06 { for 94% confidence }
Now at = 0.06 with DF = 41 , can be obtained from statistical book orfrom any software like R/Excel
From R
>qt(1-0.06/2,df=41) [1] 1.934283
Hence = 1.934283
Thus
CI = { – * , + * }
CI = { 132.5024 – 1.934283 * 
, 132.5024 + 1.934283 * }
CI = { 122.4178 , 142.587} [ after calculation ]
Hence 94% two-bounds confidence interval on the diameteris { 122.4178 , 142.587}
b) Construct a 94% lower confidence bounds on the diameter.
94% lower confidence bound LB is given by
LB = – 
*
Now is t-distributed at = 0.06 with DF = 41
Calculation from R , since in can’t find accurate value at df=41 and = 0.06 in statistical book
From R
> qt(1-0.06,df=41)[1] 1.587865
Hence = 1.587865
Thus
LB = – *
LB = 132.5024 – 1.587865 *
LB = { 132.5024 – 8.278494
LB = 124.2239
{ CI = { 124.2239 , Inf } or 
> 124.2239 }
Hence 94% lower confidence bounds on the diameter is LB= 124.2239
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