Write an equivalent expression for 5n +4y + 2n
2 answers:
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Answer:
The 't' test statistic = 1.46 < 1.69
The test of hypothesis is H 0 :μ = 800 is accepted
A sample of 30 bulbs are found came from average µ= 800
Step-by-step explanation:
Step 1:-
Given population of mean μ = 800
given size of small sample n =30
sample standard deviation 'S' = 45
Mean value of the sample χ = 788
Null hypothesis µ =800
alternative hypothesis µ ≠ 800
<u>Step 2</u>:-
The 't' test statistic t =
t = = 1.4607
<u>Step 3</u>:-
The degrees of freedom γ = n-1 = 30-1 =29
From "t" value from table at 0.05 level of significance ( t = 1.69)
The calculated value t = 1.4607 < 1.69
Therefore The null hypothesis H_{0} ' is accepted.
<u>conclusion</u>:-
A sample of 30 bulbs are came found from average µ= 800