Ok, I'm a little lost in your answer. I would say:
H0:p=.02, Ha:p>.02
The expected mean number of damages is p*n = .02*60 = 1.2
The formula for the standard deviation for a binominal distribution is sqrt(n*p*q) (where q=1-p)
so, sd = sqrt(60*.02*.98) = 1.0844
the alpha for 95% confidence interval (one tailed) is 1.645. We will reject the null hypothesis if we number of damaged washers is greater than 1.2+1.645*sd = 2.98. Since we observed 5 damage washer, reject the null hypothesis.
I hope this helps.
An appliance manufacturer stockplies washers and dryers in a large warehouse for shipment to retail stores. Sometimes in handling then the appliances get damaged. Even though the damage may be minor, the company must sell those machines at drastically reduced prices. The company goal is to keep the level of damaged machines below 2%. One day an inspector randomly checks 60 washers and finds that 5 of them have scratches or dents. Is this strong evidence that the warehouse is failing to meet the company goal? Test an appropriate hypothesis and state you conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.
p= proportion of washes that are damaged
H0:p=.002 Ha:p>.002
Plausible Independence: assume washes are independent of each other
success/failure (.002)(.60)=12>10
10% condition: there are at least 600 washers
z=.0834-.05 (divided by)/square root (.002)(.098) (divided by) 60
equals=.00122
okay now that is how much I have been able to figure out but I don't know if all of that is correct and I need to figure out the p-value and a conclusion. please help?
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