Answer: There is sufficient evidence to reject the dealer's claim that the mean price is at least $20,500
Step-by-step explanation:
given that;
n = 14
mean Ж = 19,850
standard deviation S = 1,084
degree of freedom df = n - 1 = ( 14 -1 ) = 13
H₀ : ц ≥ 20,500
H₁ : ц < 20,500
Now the test statistics
t = (Ж - ц) / ( s/√n)
t = ( 19850 - 20500) / ( 1084/√14)
t = -2.244
we know that our degree of freedom df = 13
from the table, the area under the t-distribution of the left of (t=-2.244) and for (df=13) is 0.0215
so P = 0.0215
significance ∝ = 0.05
we can confidently say that since our p value is less than the significance level, we reject the null hypothesis ( H₀ : ц ≥ 20,500 )
There is sufficient evidence to reject the dealer's claim that the mean price is at least $20,500
Answer:
99.7%
Step-by-step explanation:
Using the z score formula
z-score is z = (x-μ)/σ
where:
x = raw score
μ = population mean
σ = population standard deviation.
a) for x = raw score = 35
μ = population mean = 44
σ = population standard deviation =3
z = (35 - 44)/3
z = -9/3
z = -3
b) for x = raw score = 53
μ = population mean = 44
σ = population standard deviation =3
z = (53 - 44)/3
z = 9/3
z = 3
We would use the standard normal distribution table to find their probabilities
P(Z<=-3)=0.0013499
P(Z<=+3)=0.99865
So P(-3<=Z<=3)=0.99865 - 0.0013499 = 0.9973
Converting to percentage = 0.9973 × 100 = 99.73%
Therefore, the percentage of auto batteries of this brand that have a life of 35 to 53 months to one decimal place is 99.7%
Multiply both sides by 2pi:
F(2pi) = sqrt(g/l)
Square both sides:
F^2 (2pi)^2 = g/l
Solve for l by dividing both sides by F^2 (2pi)^2
L = g/ F^2 (2pi)^2
Simplify:
L = g / 4pi^2(f^2)
Answer:
108
Step-by-step explanation:
