Question

A sample of 64 account balances from a credit company showed an average daily balance of $1,040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i.e., population mean) is significantly different from $1,000. 1. State the null and alternative hypothesis. 2. Compute the test statistic. 3. Compute the p-value. 4. Based on the p-value and the significance level of 0.05% make a decision regarding the hypotheses. 5. Interpret your results in the context of the problem.

Answer 1

Answer:

See step by step exlanation

Step-by-step 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 $