Asample of 1700 computer chips revealed that 51% of the chips do not fail in the first 1000 hours of their use. the company's promotional literature states that 48% of the chips do not fail in the first 1000 hours of their use. the quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage.
1. determine the decision rule for rejecting the null hypothesis, h0, at the 0.01 level.

The answer is -1.2

Step-by-step explanation:

Recalling the standard deviation formula,

we have : σ = sqrt[ P * ( 1 - P ) / n ]

Where sqrt = square root

n = Number of derivative

Therefore,:

if, H0: p is not 48%  vs H1: p1 = 48%

standard deviation is:

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt[ .48*.49/900 ] = .01666333+ z -score is (p-P)/σ

= (.51-.53)/.01666333

The answer therefore =  -1.2

 ICONCLUSION:

It can therefore be concluded that there is statistical evidence that the proportions are different. H0 accepted in the hypothesis.

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 99% confidence interval) reject Null hypothesis

Z score < Z(at 99% confidence interval) accept Null hypothesis

Null hypothesis: H0 = 0.48

Alternative hypothesis: Ha <> 0.51

z score = 2.48

P value = P(Z<-2.48) + P(Z>2.48) = 0.006569 + 0.006569= 0.013

Since z at 0.01 significance level is between -2.58 and +2.58 and the z score for the test (z = 2.48) falls with the region bounded by Z at 0.01 significance level. And also the two-tailed hypothesis P-value is 0.013 which is greater than 0.01. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 1% significance level the null hypothesis is valid.

Step-by-step explanation:

Given;

n=1700 represent the random sample taken

Null hypothesis: H0 = 0.48

Alternative hypothesis: Ha <> 0.51

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1700

po = Null hypothesized value = 0.48

p^ = Observed proportion = 0.51

Substituting the values we have

z = (0.51-0.48)/√{0.48(1-0.48)/1700}

z = 2.476

z = 2.48

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-2.48) + P(Z>2.48) = 0.006569 + 0.006569= 0.013

Since z at 0.01 significance level is between -2.58 and +2.58 and the z score for the test (z = 2.48) falls with the region bounded by Z at 0.05 significance level. And also the two-tailed hypothesis P-value is 0.013 which is greater than 0.01. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 1% significance level the null hypothesis is valid.

i got 10.28 but that rounds up to 10.3