Mean annual water consumption per household in a certain city is 6,800 L. The variance is 1,440,000. A random sample of 40 households in one neighborhood reveals a mean of 8,000. a. Test the...

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(Question a)
I Test: µ 6= 6800 with normal distn
Hypothesis test:
H0 : µ = 6800 (Null Hypothesis)
Ha : µ 6= 6800 (Alternative Hypothesis, also called H1)
This is two tailed test.
x = 8000
σ2 = 1440000 =⇒ σ = 1200
n = 40
Significance Level, α = 0.05 (If no value is given, we take level of 0.05)
Test statistic z? = x− µ
σ/
√
n
Test Statistic, z? = 8000− 6800
1200/
√
40
≈ 6.324555 ≈ 6.32
Test Statistic, z? = 6.32
P-value Approach:
P-value is in direction of Alternative hypothesis.
Since Ha : µ 6= 6800, P-value= P (Z < −6.32 or Z > 6.32)
P-value is Area on left of −6.32 and on right of test statistic = 6.32
0.0000
−6.32
0
0.0000
test statistic = 6.32
P-value = 2 ∗ P (Z < −6.32) = 2 ∗ 0 =0.0000 (from z-table)
P-value = 0.0000
Rejection Rule: Reject H0 if p-value < α
Using excel function 2*normdist( -abs( (8000 - 6800)/( 1200/sqrt(40))),0,1, TRUE) or TI’s
function 2*normalcdf( 1E-99, -abs( (8000 - 6800)/( 1200/sqrt(40))) ) answer is: 0.0000000002
Decision: Since 0.0000 < 0.05, we reject the null hypothesis.
Rejection Region Approach (Traditional Method)
The Critical Value = zcrit = zα/2 = z0.025 =1.96 (From z table)
Critical Values ±1.96
Rejection Rule: Reject H0 if |z?| >...