1600 computer chips revealed that 24% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 26% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Is there enough evidence at the 0.10 level to support the manager's claim?

There is not enough evidence to support the mangers claim.

Step-by-step explanation:

We are given;

Sample Size; n = 1600

Sample proportion; p^ = 0.26

Population Proportion; p = 0.24

The hypotheses are;

Null Hypothesis: H0: p = 0.24

Alternative Hypothesis: Ha: p ≠ 0.24

The z-formula for this is;

z = (p^ - p)/√(p(1 - p)/n)

Plugging in the relevant values, we have;

z = (0.26 - 0.24)/√(0.24(1 - 0.24)/1600)

z = 0.02/0.01067707825

z = 1.87

From online p-value from z-score calculator, with 2 tail attached, we have;

p = 0.061484

This p-value is less than the significance level of 0.1. Thus, we will reject the null hypothesis and conclude that there is not enough evidence to support the managers claim.

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.