PLEASE HELP The sample of six measurements shown below was randomly selected from a normally distributed population. Complete parts a through c.

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The following sample of six measurements was randomly selected from a normally distributed population: 1, 3, -1, 5, 1, 2.

a. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis,

μ<3  Use  α=.05

b. Test the null hypothesis that the mean of the population is 3 against the alternative hypothesis,

μ≠3  Use  α=.05

c. Find the observed significance level for each test.

Step-by-step explanation:

We will first find the standard deviation (S.D) of the six measurements.

To find the S.D, we calculate the mean and variance

Mean = ∈n/n where ∈n = sum of set of numbers n

Mean = [1+3+(-1)+5+1+2] / 6 where n = 6

Mean = 11/6 = 1.8

Variance = Sum of squared deviation from mean / n-1

Variance = [(1-1.8)² + (3-1.8)² + (-1-1.8)² + (5-1.8)² + (1-1.8)² + (2-1.8)²] / 6-1

Variance = [0.64 + 1.44 + 7.84 + 10.24 + 0.64 + 0.04] / 5 = 20.84/5 = 4.168

Standard Deviation S.D = √Variance = √4.168 = 2.041

(a) : μ=3,   : μ<3

Using  α=0.05

test stat =  {mean - μ] / S.D / = [1.8 - 3] / [2.041/ √6

test stat = -1.2 / 0.833 = -1.44

Critical Value from Student T distribution is t = 2.015

Therefore, the rejection value contains values < -2.015

-1.400 > -2.015. It Fails to reject

(b) : μ = 3,   : μ ≠ 3

Using  α=0.05

test stat =  {mean - μ] / S.D / = [1.8 - 3] / [2.041/ √6

test stat = -1.2 / 0.833 = -1.44

Critical Value from Student T distribution is t = 2.571

Therefore, the rejection value contains values < -2.571 and larger value to 2.571

2.571 > -1.400 > -2.571. It Fails to reject

(c) Significance Value of (a) is P-value > 0.100

Significance Value of (b) is P-value > (2 x 0.100 = 0.200)

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