# Final Cheat Sheet

**Topics:**Sample size, Normal distribution, Standard deviation

**Pages:**3 (628 words)

**Published:**February 6, 2013

*A credit card company wondered whether giving frequent flyer miles for every purchase would increase card usage, which has a current mean of $2500 per year. They gave free flyer miles to a simple random sample of 25 card customers and found the sample mean to be $2542 and the standard deviation to be $109. n= 25 Ho (Claim) µ=2500 OR Ha µ > 2500 *Use t-table n .50 Ha a. 5% level of significance- Reject Ho, there is enough evidence to accept claim P ≤ .50 Hob. Error? Type I

N = 4000c. Right tailed

P = 2200/4000 d. CV = 1.645 Test Statistic= 6.325

z=p-ppqn

*A sample of 54 watercraft accidents reported to Nebraska revealed that 85% of them involved personal watercrafts. Suppose the national average of watercraft accidents is 78%. Does the accident rate in Nebraska exceed the rate in the nation? Use 0.01 level of significance. n=54 p≤ 0.78 Ho (Claim)

p=.85 p> 0.78 Ha

α=.01

z=p-ppqn Test Statistic= 1.24 CV= 2.33

Minimum Sample Sizes

*You wish to estimate, with 99% confidence , the proportion of computers that need repairs or have problems by the time the product is three years of. Your estimate must be accurate within 2% of the true proportion. Find the minimum sample size with no preliminary estimate. n=(Zc* σE)^2 n=pq(ZcE)^2 CV = 2.575 C = 99% *Confidence intervals are two tails!*

E= .02 *Must be rounded up with minimum sample size* p= .50 =4,145

Binomial

*When an assembly machine is properly calibrated approximately 5% are defective. Randomly selects 20 products. Production is halted if anything is defective. What is the probability the production must be halted? p(x)=(nCx)(px)(qn-x) n=20 x=0 p=.05 q=.95 Answer: 1-.95^20...

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