The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a standard deviation of 1.0. A level of significance of 0.05 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

We should reject H₀   for values of | x₁| > 5,26     |x₁| > 6,12

Step-by-step explanation:

Sample mean    μ₀  =  5,7

Sample size       n₁  = 24

Sample standard deviation   s =  1

Significance level    α  = 0,05        CI = 95 %

Hypothesis criteria: Valves don´t meet the specification, which means pressure could be higher or lower than nominal

Normal Distribution where n < 30 we should use a two-tail t-student test

degree of freedom     df =  24 - 1  = 23

And with df = 23  and  α = 0,025 we find in t-tables the value for

t(c) =  - 2,0687     ( on the left tail )

Test Hypothesis:

Null Hypothesis                                       H₀            x  = μ₀

Alternative Hypothesis                           Hₐ            x  ≠ μ₀

To compute  t(s)

t(s) = ( x  -  μ₀ ) / s/√n

t(s) = ( x - 5,7 )*4,80

t(s) =  4,80*x  -  27,34

Then   for values of  t(s)   |t(s)| > 2,0687 we have to reject H₀

If we make   t(s)  =  t(c)  we find

-2,0687 = 4,8*x - 27,34

x  = ( 27,34 - 2,07 ) / 4,8      and on the other tail  x  = 6,12

x = 5,26

Therefore for values of 5,26 ( under )   and above 6,12 we shoud reject the Null hypothesis

45

step-by-step explanation:

to find a percentage, add a decimal in front and multiply it by the other number.

.25 times 180= 45

.19875994384

step-by-step explanation:

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