A light bulb manufacturer guarantees that the mean life of a certain type of lig

Question

A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 774

do you have enough evidence to reject the manufacturer's claim? Complete parts (a) through (e).

A light bulb maracturar guarantees that the mean Itat Cartain type at Ight hub IR at least 774 hours. A random sample of 29 light bulhs has a mean Ita of 746 hours Aaaluma ha population is normally distributed and the population standam datation is 61 hours Ata= |1 H, do you have enough eviclenca to nejach the manufacturer's clain? Complete paita (a) througin ex (a) Identify the nulhypothesis and a temative hypothesis OB. He 112 774 H.: PS774 (daim} OE. Hg. pä746 II, p746 (claim) CC. He pu*TTAidein H.:-774 CE H. pc746 (clasin II, 2746 OA. Ho = 746 Hypt 745 (caim) OD. Ho. pä774 fusim II, 774 (6] Identify the cailical value(s). Use technologue. 26-0 (Is A Fam rate AnRMATA , Td Rtl to F43 61ple:48 4.4 needed Icientify the reaction reglans) Chance the corect answer below. Fail to reject , Q Fall to reject H. (c) lidertify the standardized test statistic. Use technology 2=C(Round to two decimal places 58 naate:1 ] (d) Decide whether to reject or fail to reject the null hypothesis, and ici interpret the decision in the context of the original claim. OA. Reject Ho. There is sufficient evidence to reject the claim that mean bulb life is at least 774 hours. OC. Fail po reject Hy. There is sufficient evidence to reject the claim that mean bulb life is at least 774 hours OB. Reject Ho. There is not sufficienti evidence lo reject, die ciain disel imeon bulb ife is at least 774 hours OD. Fail to reject He. There is not suficient. cvidence to reject the claim that mean tub life is at least 774 hours.

Explanation / Answer

Given that,
population mean(u)=774
standard deviation, =61
sample mean, x =746
number (n)=29
null, Ho: >=774
alternate, H1: <774
level of significance, = 0.08
from standard normal table,left tailed z /2 =1.405
since our test is left-tailed
reject Ho, if zo < -1.405
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 746-774/(61/sqrt(29)
zo = -2.47
| zo | = 2.47
critical value
the value of |z | at los 8% is 1.405
we got |zo| =2.47 & | z | = 1.405
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -2.47 ) = 0.01
hence value of p0.08 > 0.01, here we reject Ho
ANSWERS
---------------
null, Ho: >=774
alternate, H1: <774
critical value: -1.405
option c
test statistic: -2.47
decision: reject Ho, there is sufficent evidence to reject the claim that the mean bulb life is atleast 774
p-value: 0.01

we dont have evidence to support teh claim

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