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.

at first, in 1635, the evening meetings had been just for women . .

what relationship between ideas is shown in the text above?

question 1 options:

location of events

similarity of events

timing of events

examples of an idea

step-by-step explanation:

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05