1.) The life in hours of a battery is known to be approximately normally distributed with a standard deviation of σ = 1.25 hours. A random sample of 10 batteries has a mean life of of x̄ = 40.5 hours.

A.) Is there evidence to support the claim that battery life will exceed 40 hours? use α=0.05

Ho : u ≤ 40
Ha : u > 40

X = 40.5   n = 10    σ = 1.25    α = 0.05   df = 9

Z = x̄ - μ
    --------
     σ / √n


z = -1.264911           α = 0.05 => 1.83

-1.264911 < 1.83

Comment: With an alpha level of 0.05 there was not enough evidence to support that the battery’s life will not exceed 40 hours. Hence we failed to reject Ho.

B.) What is the P-Value for the test in part A?

1.83 => 41.22,

p( Z > 1.83) = 1 - P(Z < 1.83)            

             = 1 - 0.95
           
p( Z > 1.83) = 0.05

p( Z > -1.264911) = 1 - P(Z < -1.264911)            

             = 1 - 0.1029516
           
p( Z > -1.264911) = 0.8970484

0.8970484 > 0.05

C.) What is the β-error for the text in part B if the true mean life is 42 hours?

β: P(X < 41.22 | μ = 42)

Z of 41.22 to 42 => -1.973261

P(Z < -1.973261) = 0.0242329

Comment: the probability of having a β-error if the true mean life is 42 hours is very low at only 2.42329%.

D.) What sample size would be required to ensure that β does not exceed 0.10 if the true mean is is 44 hours?

β: P(X < 41.22 | μ = 44) = 0.10

n = (Za + Zb)^2 * σ^2
    -----------------
           δ^2

n = 1.233686

Comment: If the true mean is 44 no matter how many the sample size is used the chances of having β-error will never exceed 0.10. As seen with the equation is that to have at least 1.233686, which would just be rounded off to 1, as a sample size to only have 0.10 chance for a β-error. If you would increase your sample size β-error will get smaller.

E.) Explain how you could answer the question in part A by calculating an appropriate confidence bound on battery life

It is important to set a proper confidence bound so that you can minimize having an α-error and a β-error. One way that you would be able to reject Ho in part A would be changing it from a right tail to a left tail and decrease the confidence bound 0.90 which would have a z-score of -1.28.

2.) Brand A gasoline was used in 16 similar automobiles under identical conditions. The corresponding sample of 16 values (miles per gallon) had mean 19.6 and standard deviation 0.4. Under the same conditions, high-power brand B gasoline gave a sample of 16 values with mean 20.2 and standard deviation 0.6. Is the mileage of B significantly better than that of A? Assume normality. Test the hypothesis using both P-value and fixed significance level with a = 0.05 approaches (if possible).

Ho: u1 = u2
Ha: u1 < u2

Brand A

u = 19.6   n = 16
σ = 0.4    α = 0.05
df = 15

Brand B

u = 20.2   n = 16
σ = 0.6    α = 0.05
df = 15

Solution

Zo = u1 - u2
    ---------
   σ1^2   σ2^2
 √(---- + ----)
    n1     n2

Zo = -3.328201

P(Z < -3.328201) = 0.0004370436

0.0004370436 < 0.05

Comment: With a significance level of 0.05 there is enough evidence to say that brand B is significantly better than that of brand A. In this scenario we reject the null hypothesis