An engineer selected 30 components at random and measure their strengths. 34,54,73,38,89,52,75,33,50,39,42,42,40,66,72,85,28,71,52,47,41,36,33,38,49,51,55,63,72,78 Does this data suggest that the population mean strength of the components differs from 50? Set up an appropriate hypothesis test to answer this question.

In the given question, an engineer selected 30 components at random and measure their strengths. 34,54,73,38,89,52,75,33,50,39,42,42,40,66,72,85,28,71,52,47,41,36,33,38,49,51,55,63,72,78

We have to check does this data suggest that the population mean strength of the components differs from 50 and we have to set up an appropriate hypothesis test to answer this question.

Sample size n = 30

Claim : The population mean strength of the components differs from 50.

Null and alternative hypothesis:

H(0): μ = 50

H(A): μ≠50

As the population standard deviation is unknown (not given) , t-test can be used .

Test Statistic :

t = (X-μ)/(S/√n)

To find the sample mean we find the average of the given sample.

So

X = 34 + 54 + 73 + 38 + 89 + 52 + 75 + 33 + 50 + 39 + 42 + 42 + 40 + 66 + 72 + 85 + 28 + 71 + 52 + 47 + 41 + 36 + 33 + 38 + 49 + 51 + 55 + 63 + 72 + 78/ 30

X = 53.267

Now finding the standard deviation using the excel.

SD = 17.1101

To find the SD we use the formula =STDEV(B1:C15)

Now putting the value in the formula of Test Statistic

t = (X-μ)/(S/√n)

t = (53.267-50)/(17.1101/√30)

After solving:

t=1.046

P-value :

As the alternative hypothesis H(A) is of not equal to type , this is two tailed test.

Degrees of freedom = n-1 = 30 -1 = 29

df =29

Using Excel function , =TDIST( t , df , number of tails )

=TDIST( 1.046 , 29 , 2 )

=0.304201657

Rounding to three decimal places

= 0.304

P-value = 0.304

Hence, the population mean strength of the components differs from 50 is 0.304.

To learn more about Test Statistic link is here

brainly.com/question/15236063

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