You are the engineer in charge of evaluating a new product for potential use in a hydraulic system. The manufacturer has sent 100 randomly chosen samples for you to test. You plan to go through with the deal as long as you can prove that less than 8% of the products you subsequently purchase will be defective. The engineering and business teams at your company believe a 90% confidence level is appropriate for the test.

(a) Specify the null and alternative hypotheses for your test.
(b) Describe (in words) a Type I error for this deal.
(c) Out of the 100 samples provided by the manufacturer, at most how many can be defective for you to agree to use the new product?
(d) For better or worse, your boss convinces you to go through with the deal. Turns out the minimum order is 5000 pieces. Assuming you purchase that many pieces of the new product, and that you found 6 defective pieces out of the 100, generate a 90% two-sided confidence interval for the number of pieces that will be flawed.

a) The null hypothesis is represented as

H₀: p ≥ 0.08

The alternative hypothesis is represented as

Hₐ: p < 0.08

b) A type I error for this question would be that

we conclude that the proportion of defective products is less than 8% when in reality, the proportion of defective products, is more than or equal to 8%.

c) At most, the number of defective products in the sample for you to agree to use the new product = 7

d) If minimum of 5000 pieces are purchased, 90% confidence interval for minimum number of flawed pieces will be (103, 497)

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and is usually stating the opposite of the theory is being tested. It usually maintains that random chance is responsible for the outcome or results of any experimental study/hypothesis testing. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared. It usually confirms the the theory being tested by the experimental setup. It usually maintains that other than random chance, there are significant factors affecting the outcome or results of the experimental study/hypothesis testing. It usually contains the signs ≠, < and > depending on the directions of the test

For this question, we want to prove that less than 8% of the products we subsequently purchase will be defective.

So, the null hypothesis will be that there is not enough evidence in the sample to say that less than 8% of the products we subsequently purchase will be defective. That is, the proportion of the sample that are defective is more than or equal to 8%.

And the alternative hypothesis is that there is enough evidence in the sample to say that less than 8% of the products we subsequently purchase will be defective.

Mathematically,

The null hypothesis is represented as

H₀: p ≥ 0.08

The alternative hypothesis is represented as

Hₐ: p < 0.08

b) A type I error involves rejecting the null hypothesis and accepting the alternative hypothesis when in reality, the null hypothesis is true. It involves saying that there is enough evidence in the sample to say that less than 8% of the products we subsequently purchase will be defective when in reality, there isn't enough evidence to arrive at this conclusion.

That is, the proportion of defective products in reality, is more than or equal to 8% and we have concluded that the proportion is less than 8%.

c) Out of the 100 samples provided by the manufacturer, at most how many can be defective for you to agree to use the new product?

The engineer agrees to use the new product when less than 8% of the products we subsequently purchase will be defective.

8% of the product = 0.08 × 100 = 8.

Meaning that the engineer agrees to subsequently purchase the product if less than 8 out of 100 are defective.

So, the maximum number of defective product in the sample that will still let the engineer purchase the products will be 7.

(d) For better or worse, your boss convinces you to go through with the deal. Turns out the minimum order is 5000 pieces. Assuming you purchase that many pieces of the new product, and that you found 6 defective pieces out of the 100, generate a 90% two-sided confidence interval for the number of pieces that will be flawed.

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = 0.495

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 90% confidence interval for sample size of 100 using the t-tables since information on the population standard deviation.

Degree of freedom = n - 1 = 100 - 1 = 99

Significance level = (100-90)/2 = 5% = 0.05

Critical value = t(0.05, 99) = 1.660

Standard error of the mean = σₓ = √[p(1-p)/n]

p = 0.06

n = sample size = 100

σₓ = (0.06/√100) = 0.006

σₓ = √[0.06(0.94)/100] = 0.0237486842 = 0.02375

90% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.06 ± (1.660 × 0.02375)

CI = 0.06 ± 0.039425

90% CI = (0.020575, 0.099425)

90% Confidence interval = (0.0206, 0.0994)

If minimum of 5000 pieces are purchased, 90% confidence interval for minimum number of flawed pieces will be

5000 × (0.0206, 0.0994) = (103, 497)

Hope this Helps!!!

answer: 20

step-by-step explanation: subtract the minimum date value from the maximum data value to find the date range. in this casem the date range is 42 - 22 = 20

hope this ! : ) ~zane