Answer:
H0 rejected as μ>4
and a fish is unsafe to eat from the lake.
Step-by-step explanation:
1)Let the null hypothesis be H0 : mu ≤4 against the alternate Ha: mu > 4 ppb
H0: μ ≤4 against the claim Ha: μ>4
2) X`= ∑x/ n= 2.9 +7.6+ 4.8+ 5.2+ 5.1+ 4.7+ 6.9+ 4.9+ 3.7+ 3.8/10
= 49.6/10= 4.96
The standard deviation can be calculated sigma= 1.344767 ( using statistic calculator)
3) The significance level is taken to be ∝=0.05
The value of z at 0.05 for 1 sided test is z >± 1.645
i.e the critical region is less than - 1.645 and greater than +1.645
4) Taking the distribution to be approximately normal
Z= x`- u / s/ √n
Z= 4.96-4/ 1.345/10
Z= 4.96-4/ 1.345/3.1622
Z= 0.96/0.4253
Z= 2.257
5) Since the calculated value of z = 2.257 falls in the critical region we reject our null hypothesis and conclude that true mean value of the PCB concentration is greater than 4 (ppb) and that a fish is unsafe to eat from the lake.
Answer:
first you have to move the varible so
5x +10=5y
then thats as far as i can go
good luck
Answer:
90
45 x 200
100
= 90
Step-by-step explanation:
