Claim: Most adults would not erase all of their personal information online if they could. A software firm survey of 669 randoml
1 answer:
Answer:
a) p=0.39, where p the parameter of interest represent the true proportion of adults that would erase all their personal information online if they could
b) Null hypothesis:
Alternative hypothesis:
Step-by-step explanation:
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
On this case the claim that they want to test is: "The true proportion of adults that would erase all their personal information online if they could is 0.39 or 39%". So we want to check if the population proportion is different from 0.39 or 0.39%, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.
Part a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information.
p=0.39, where p the parameter of interest represent the true proportion of adults that would erase all their personal information online if they could
Part b. Identify the null and alternative hypotheses.
Null hypothesis:
And for the alternative hypothesis we have
Alternative hypothesis:
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Answer:
a) P (0 defective component) = 0.5
P( 1 defective component ) = 0.35
P( 2 defective component ) = 0.15
b) P( 0 ) = 0
p ( 1 ) = 1
p ( 2 ) = 0.33
Step-by-step explanation:
p( no defective ) = 0.5
p( 1 is defective ) = 0.35
p( 2 is defective ) = 0.15
Given that 2 components are selected at random
<u>a) Given that neither component is defective </u>
Probability of 0 defective component = 0.5
P( 1 defective component ) = 0.35
P( 2 defective component ) = 0.15
<u>b) Given that one of the two tested component is defective </u>
P( 0 defective ) = 0
P( 1 defective ) = P ( ) = p( x = 1 ) / 1 - P ( x = 0 )
= ( 0.5 )^1 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^1
= 0.5 * 1 / 1 - 0.5 = 0.5 / 0.5 = 1
p ( 2 defective ) = p( x = 3 ) / 1 - P ( x = 0 )
= ( 0.5 )^2 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^2
= 0.25 / 0.75 = 0.33
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