written 7.7 years ago by
teamques10
★ 64k
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modified 7.7 years ago
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n =50 ($\boldsymbol{\gt}$30 ,so it is large sample)}
$$\boldsymbol{\alpha }\boldsymbol{=}\boldsymbol{14}.\boldsymbol{5}\boldsymbol{\ ,\ }\boldsymbol{\sigma }\boldsymbol{=}\boldsymbol{2}.\boldsymbol{2}$$
Step 1 :
Null hypothesis ($H_0):\ \mu =15.6\ (i.e.\ the\ sample\ belongs\ to\ the\ population)$
Alternative hypothesis : $\mu \neq 15.6\ (i.e.\ the\ sample\ does\ not\ belong\ to\ the\ population)$
(Two failed tests)
Step 2:
L.O.S = 5%
$\mathrm{\therefore }$ Critical value ($z_{\alpha })=19.6$
Step 3 :
Since the sample is large,
S.E. = $\sigma /\sqrt{n}=2.2/\sqrt{50}$ = 0.3111
Step 4 :
Test statistic :
$$z_{cal} = 14.5 -- 15.6/0.3111 = -3.5355$$
Step 5 :
Decision : Since |$z_{cal}|\gt z_{\alpha }$ , $H_0\ $is rejected.
Thus sample is not from a batch having a mean breaking strength of 15.6 pound.