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3 years ago

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A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week

. Of the 250 employed individuals surveyed,42 responded that they did work at home at least once per week. Construct a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.
The lower bound is nothing.

(Round to three decimal places as needed.)

1 answer:

Travka [436]3 years ago

4 0

Answer:

solution is given below

Step-by-step explanation:

A simple random sample size n = 250 . Of the 250 employed individuals surveyed,42 responded that they did work at home at least once per week.

Construct 99% confidence interval for population

For proportion : 42 / 250 = 1/10 = 0.16

Mean = 2.5 * sqrt [ 0.1 * 0.9 / 250]

= 2.5 * 0.01

= 0.47

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

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(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

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Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

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Question 10: Who is likely to get number 5

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From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

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<em>From the above calculations: </em> $0.267 > 0.03125$ <em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

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This means that the median is the mean of the 16th and the 17th item

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$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

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This means that the median is the mean of the 15th and the 16th item. So,

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<em>Person B has a greater median of 5</em>

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From the sample space of person B, we have:

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Question 13: Person with higher probability of 3 or more

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