A plastics manufacturer has developed a new type of plastic trash can and proposes to sell them with an unconditional 6-year warranty. To see whether this is economically feasible, 20 prototype cans are subjected to an accelerated life test to simulate 6 years of use. The proposed warranty will be modified only if the sample data strongly suggests that fewer than 90% of such cans would survive the accelerated test. After the accelerated life test on the 20 prototype cans, 13 survive. Does this indicate that the proposed warranty should be modified? Use =0.05.

since -3.73 is less than 1.645, we reject H₀.    

Therefore this indicate that the proposed warranty should be modified

Step-by-step explanation:

Given that the data in the question;

p" = 13/20 = 0.65

Now  the test hypothesis;

H₀ : p = 0.9

Hₐ : p < 0.9

Now lets determine the test statistic;

Z = (p" - p ) / √[p×(1-p)/n]

= (0.65 - 0.9) /√[0.9 × (1 - 0.9) / 20]  

= -0.25 / √[0.9 × 0.1 / 20 ]

= -0.25 / √0.0045

= -0.25 / 0.067

= - 3.73

Now given that a = 0.05,

the critical value is Z(0.05) = 1.645 (form standard normal table)

Now since -3.73 is less than 1.645, we reject H₀.    

Therefore this indicate that the proposed warranty should be modified

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

yes the data was not prsented

step-by-step explanation: