# (a) The company's production equipment produces metal discs weighing 200 g. It should be noted that the weight of the discs corresponds to the normal

(a) The companys
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a)
The formula for estimation is:
$\mu =M±Z\left({s}_{M}\right)where:$
$M=$ sample mean
$Z=Z$ statistic determined by confidence level
${s}_{M}=\text{standard error}=\sqrt{\left({s}^{2}\text{/}n\right)}$
Step 2
Calculation
$M=205$
$Z=2.58$
${s}_{M}=\sqrt{\left({7}^{2}\text{/}20\right)}=1.57$
$\mu =M±Z\left({s}_{M}\right)$
$\mu =205±2.58\cdot 1.57$
$\mu =205±4.03$
Result
$M=205,99\mathrm{%}CI\left[200.97,209.03\right]$
Part b
Z Score Calculations
$Z=\left(M-\mu \right)\text{/}\sqrt{\left({\sigma }^{2}\text{/}n\right)}$
$Z=\left(295-300\right)\text{/}\sqrt{\left(8\text{/}20\right)}$
$Z=-5\text{/}0.63246$
$Z=-7.90569$
The value of z is -7.90569. The value of p is $<.00001$. The result is significant at p $<.05$