Suppose the average lifetime of a certain type of car battery is known to be 60 months. Consider conducting a two-sided test on it based on a sample of size 25 from a normal distribution with a population standard deviation of 4 months. a) If the true average lifetime is 62 months and ďˇ =0.01, what is the probability of a type II error

0.53187998

Step-by-step explanation:

We are given;

Population mean; μ = 60

Sample size; n = 25

Population standard deviation; σ = 4

Significance level; α = 0.01

Let's define the hypothesis;

Null hypothesis;H0: μ = 60

Alternative hypothesis;Ha: μ ≠ 60

Z-score for a significance level of 0.01 from tables is z = 2.575

Since it's a 2 sided test, then;

Z-score at lower limit = -2.575

Z-score at upper limit = 2.575

We know that z-score formula is;

z = (x¯ - μ)/(σ/√n)

At lower limit;

-2.575 = (x¯ - 60)/(4/√25)

-2.575 × 0.8 = x¯ - 60

-2.06 = x¯ - 60

x¯ = 60 - 2.06

x¯ = 57.94

At upper-limit;

2.575 = (x¯ - 60)/(4/√25)

2.575 × 0.8 = x¯ - 60

2.06 = x¯ - 60

x¯ = 60 + 2.06

x¯ = 62.06

the probability of a type II error is expressed as:

β = P(57.94 < x¯ < 62.06)

If the true average lifetime is 62 months, then;

β = P((57.94 - 62)/(4/√25)) < z < (62.06 - 62)/(4/√25)

β = P(-5.08 < z < 0.08)

β = P(Z < 0.08) - P(Z < -5.08)

From z-distribution table attached, we have; P(Z < 0.08) = 0.53188

Also, from z-distribution calculator online, we have;

P(Z < -5.08) = 0.0000000189

Thus;

β = 0.53188 - 0.0000000189

β ≈ 0.53187998

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