The sample observations on stabilized viscosity of asphalt specimens are 2781, 2900, 3013, 2856 and 2888. Suppose that for a particular application, it is required that true average viscosity be 3000. Does this requirement appear to have been satisfied? State and test the appropriate hypothesis at α = 0.05.

there is no sufficient evidence to prove that viscosity is not 3000

Step-by-step explanation:

Let's state the hypothesis.

Null hypothesis; H0: μ = 3000

Alternative hypothesis; μ ≠ 3000

The sample mean is;

x = (2781 + 2900 + 3013 + 2856 + 2888)/5 = 2887.6

Standard deviation σ = √variance = √[(2781 - 2887.6)² + (2900 - 2887.6)² + (3013 - 2887.6)² + (2856 - 2887.6)² + (2888 - 2887.6)²)(1/(5 - 1))]

σ = √((1/4)(11363.56 + 153.76 + 15725.16 + 998.56 + 0.16)

σ = 84.026

Formula for test statistic is given as;

t = (x - μ)/σ

t = (2887.6 - 3000)/84.026

t = -1.338

From online p-value calculator from t value attached, using DF = 5 - 1 = 4, two tail and a significance level of 0.05,we have;

The p-value is 0.251896.

This is greater than the significance level of 0.05.

Thus, we will fail to reject the null hypothesis and conclude that there is no sufficient evidence to prove that viscosity is not 3000

.25

step-by-step explanation:

eve and marks ratios are equivalent

step-by-step explanation: