Answer:
See step by step exlanation
Explanation:
Assuming normal distribution
N( μ₀ , σ )
In this case N ( 1000, 200)
From a random sample of 64 accounts we get:
n = 64
x = sample mean x = 1040
σ = standard deviation of the population σ = 200
1.-Test Hypothesis
Null Hypothesis H₀ x = μ₀ = 1000
Alternative Hypothesis Hₐ x ≠ μ₀
The alternative hypothesis implies a two tail-test. Given that n > 30 and σ is known we will use z-table
2.- z(s) = ( x - μ₀ ) / σ/√n
z(s) = ( 1040 - 1000 ) / 200/ √64
z(s) = 40*8/200
z(s) = 1,6
3.-P-val for z (score 1,6) p-value = 0,0548
4.- If significance level is 0,05 then α = 0,05 and α/2 = 0,025
p-value > α/2
Then the z(s) < z(c) 1,6 < 1,96 where z(c) is z score for α/2
We should accept H₀
5.-We can claim that at 95 % (of Confidence Interval) the population mean is not significantly different from 1000 $