Question #165188

A manufacturer of electric bulbs claims that his bulbs have a mean life of 25 months with a standard deviation of 5 months. A random sample of 6 such bulbs were taken and their lifespan were recorded.

Life of​ bulbs(in months):23 25 30 20 20 12

Is the​ manufacturer's claim valid at​ 1% level of significance when the table value of t test is​ ±4.032?

​(a) Formulate a null and alternative hypothesis.

​(b) Calculate the​ mean, standard​ deviation, and the tSTAT.

​(c) Interpret the​ manufacturer's claim.


I answered a and b but c made me confused. can anyone help me?


the choices ;

A.

The null hypothesis is rejected as tSTAT is less than the table value. Hence the​ manufacturer's claim is not valid at​ 1% level of significance.

B.

The null hypothesis is not rejected as tSTAT is more than the table value. Hence the​ manufacturer's claim is valid at​ 1% level of significance.

C.

The null hypothesis is not rejected as tSTAT is less than the table value. Hence the​ manufacturer's claim is valid at​ 1% level of significance.

D.

The null hypothesis is rejected as tSTAT is more than the table value. Hence the​ manufacturer's claim is not valid at​ 1% level of significance.


1
Expert's answer
2021-02-24T07:47:30-0500

Solution

a).


H0:μ=25 vs H1:μ25H_0 : \mu =25 \space vs \space H_1 : \mu \not= 25

b). Mean, xˉ\bar{x}


xˉ=23+25+30+20+20+126\bar{x} ={23+25+30+20+20+12 \over 6}

xˉ=21.6667\bar {x} =21.6667

Standard deviation, SS


S=(xxˉ)2n1S= \sqrt {\sum (x- \bar{x}) ^2 \over n-1}

=181.333335=6.02218=\sqrt {181.33333 \over 5} =6.02218

tSTAT

tSTAT=xˉμσn=21.666672556tSTAT= {\bar{x} - \mu \over {\sigma \over \sqrt {n}}} ={21.66667 - 25 \over {5 \over \sqrt {6}}}=1.63299=-1.63299

c). Conclusion :

The null hypothesis is not rejected as tSTAT is less than the table value. Hence the​ manufacturer's claim is valid at​ 1% level of significance



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