Question

(2) yn Corporation manufactures computers chips. The machine that is used to make this chips is known to p than 4% defective chips, it needs an adjustment. The quality control inspector often selects sam of chips and inspects them for being good or defective. One such random sample of 200 chips taken recent ly from the production line contained 14 defective chips at the α-5% level of signif- cance,test whether or not the data give sufficient evidence to warrant an adjustment of the machine? Step 1. Identify the given information using proper mathematics notations: The sample size The sample proportion The significance level Are they 5? Step 2. Formulate the null and alternative hypothesis: Ho: Hi: Determine the critical value za using the significance level a-5% and using a bell-shaped diagram to tell whether this is a left-, right- or two-tailed tests: Step 3 Step 4. Assuming that the null hypothesis is true, determine the mean and standard deviation ap of the sample statistic p Step 5. Find the values of zp and the P-value: Step 6. Make your conclusion and then answer the question in common terms:
IT. Problem Solving: Show All Your Steps. At most 2 points are given to Ull work shown. with a mean (1) for wom en aged 18- 24, systolic blood pressures (in mm Hg) are normally distributed n is commonly defined as a systolic blood of 114.8 and a standard deviation of 13.1. Hypertensio pressure above 140. (a) If a woman between ages of 18 and 24 is selected at random. find the probability that her systolic blood pressure is greater than 140 (b) If a sample of 25 women in that age bracket is selected at random, find the probability that the sample mean systolic blood pressure is greater than 140 (c) If a physician has given a special treatment to a random sample of 25 women in that age bracket and find that their mean systolic blood pressure is below 140, can she conclude that none of the women have hypertension and so her treatment is 100% effective. If not, help her to modify her conclusion
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Answer #1

n = 200, np = 200 * 0.04 = 8
pcap = 14/200 = 0.07
nq = 192
alpha = 0.05

H0: p = 0.04
Ha: p > 0.04

right tailed test, z(alpha) = 1.645

mu(p) = 0.04
sigma(p) = sqrt(0.04*0.96/200) = 0.0139

test statistic,
z = (0.07 - 0.04)/0.0139 = 2.1583

p-value = 0.0155

Reject H0
There are sufficient evidence to support the claim

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