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A lumber company is making boards that are 2857.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A sample of 16 is made, and it is found that they have a mean of 2858.7 millimeters with a variance of 100.00. A level of significance of 0.05 will be used to determine if the boards are either too long or too short. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the boards are either too long or too short

Answer :

Answer:

The test statistic 't' = 0.68 < 2.1314 at 0.05 level of significance

Null Hypothesis is accepted

There is no  sufficient evidence to support the claim that the boards are either too long or too short

Step-by-step explanation:

Step(i):-

Mean of the Population (μ) = 2857.0 millimetres

size of the sample 'n' = 16

Mean of the sample (x⁻) = 2858.7 millimetres

The Variance of the sample (S²)= 100.00

Standard deviation of the sample (S) = √100 =10

Degrees of freedom = n-1 =16-1 =15

Level of significance =0.05

Tabulated value = [tex]t_{\frac{0.05}{2} , 15} = t_{0.025,15} = 2.1314[/tex]

Step(ii):-

Null Hypothesis :H₀ : There is no  sufficient evidence to support the claim that the boards are either too long or too short

Alternative Hypothesis: H₁:There is   sufficient evidence to support the claim that the boards are either too long or too short

Test statistic

                   [tex]t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }[/tex]

                  [tex]t = \frac{2858.7 -2857}{\frac{10}{\sqrt{16} } }[/tex]

                 t = 0.68

The calculated value = 0.68 < 2.1314 at 0.05 level of significance

Null Hypothesis is accepted

Final answer:-

There is no  sufficient evidence to support the claim that the boards are either too long or too short

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