5-10(|0.8-3/6|)
5-10(|1/5|)
5-(10)(0.2)
5-2
=3, so C would be your best answer.
Answer:
1/2, 9/16, 5/8, 3/4
Step-by-step explanation:
Answer:
Since the calculated value of z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.
Step-by-step explanation:
The data given is
Population mean μ= $ 7.18
Population variance= σ²= 3.81
Population Standard Deviation = √σ²= √3.81= 1.952
Sample Mean= x`= $ 8.02
Sample Standard Deviation =s = $ 2.08
Sample Size = 15
Significance Level = ∝= 0.05
The null and alternate hypotheses are
H0: σ1=σ2 against the claim that Ha: σ1≠ σ2
where σ1 is the population variance and
σ2 is the sample variance
The rejection region is Z ≥ ±1.96 for two tailed test at ∝= 0.05
The test statistic z is used
z= x`- μ/ σ/√n
Putting the values
Z= 8.02-7.18/1.952/√15
z= 0.84/0.51599
z= 1.6279
Since the calculated value of z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.
Quotient: 164
Remainder: 8/19
