10) A new process for producing synthetic diamonds can be operated at a profitab

Question

10) A new process for producing synthetic diamonds can be operated at a profitable level only if the average weight of the diamonds is greater than 5 kaat. To evaluate the profitability of the process six diamonds are generated, with recorded weights 46, .61, .52, 48, 57, and.54 karat with X-.53 and-0559 NotAssme that the relevant population distribution is normal a) Find a point estimate for the average weight of the diamonds produced by the process b) Find a point estimate of . c Is there sufficient evidence to indicate that the average weight of the diamonds produced by the rocess is greater than .5 karat? Take -0.05 (Use rejection region/critical value approach) H, :d» 0.015 at =0.10 d) Test againstA

Explanation / Answer

Given that,
population mean(u)=0.5
point of estiame for mean = sample mean, x =0.53
point of estimate for sd = standard deviation, s =0.0559
number (n)=6
null, Ho: =0.5
alternate, H1: >0.5
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.015
since our test is right-tailed
reject Ho, if to > 2.015
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =0.53-0.5/(0.0559/sqrt(6))
to =1.3146
| to | =1.3146
critical value
the value of |t | with n-1 = 5 d.f is 2.015
we got |to| =1.3146 & | t | =2.015
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.3146 ) = 0.12286
hence value of p0.05 < 0.12286,here we do not reject Ho
ANSWERS
---------------
null, Ho: =0.5
alternate, H1: >0.5
test statistic: 1.3146
critical value: 2.015
decision: do not reject Ho
p-value: 0.12286

PART C.
average rate diamond produced by process is equal to 0.50. no evidence to
support it is greater than 0.50

PART D.
Given that,
population variance (^2) =0.015
sample size (n) = 6
sample variance (s^2)=0.00312481
null, Ho: ^2 =0.015
alternate, H1 : ^2 !=0.015
level of significance, = 0.1
from standard normal table, two tailed ^2 /2 =9.236
since our test is two-tailed
reject Ho, if ^2 o < - OR if ^2 o > 9.236
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(6 - 1 ) * 0.00312481 / 0.015 = 5*0.00312481/0.015 = 1.04
| ^2 cal | =1.04
critical value
the value of |^2 | at los 0.1 with d.f (n-1)=5 is 9.236
we got | ^2| =1.04 & | ^2 | =9.236
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.9593
ANSWERS
---------------
null, Ho: ^2 =0.015
alternate, H1 : ^2 !=0.015
test statistic: 1.04
critical value: -9.236 , 9.236
p-value:0.9593
decision: do not reject Ho

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