Given that a sample of 1500 computer chips revealed that 55% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 53% of the chips do not fail in the first 1000 hours of their use.

$H_0: p = 0.53\\\\H_a: p \neq 0.53$

(Two tailed test for testing sample proportion with population proportion)

Std error of proportion assuming p = 0.53 is

$\sqrt{0.53*0.47/1000} \\=0.0157$

p difference = 0.02

Z= test statistic = p diff/std error = 1.2775~1.28

8 0

3 years ago

Solve using elimination.<br><br> -7x – 5y = 5<br> -9x – 5y = 15

ozzi

-7x-5y=5
-(-9x-5y)=15 It's minus because negative times negative = positive
2x=-10
x= -5

-7(-5) - 5y = 5
35-5y =5
-5y = -30
y=6

3 0

3 years ago

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