A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, : p 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. =  ~ N(0,1)

where, = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, test statistics =

                              = 2.491

Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)

what are the answer chices