Mathematics, 14.04.2020 23:44 allenpaietonp9v8sv

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, 41 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.

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Answer from: victo17

99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

Step-by-step explanation:

We are given that 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, 41 responded that they did work at home at least once per week.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of individuals who work at home at least once per week = $\frac{41}{250}$ = 0.164

n = sample of individuals surveyed = 250

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } }$ , $0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } }$ ]

= [0.104 , 0.224]

Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

Answer from: Quest

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90

Can someone plz answer this last ! i will give brainliest to whoever answers it

Answer from: Quest

Answercounterclockwise rotation about the origin by 180 degrees followed by reflection about the y-axisexplanationthe given figure has vertices $p(1,2),q(2,1),r(3,2),s(3,3)$ recall that for counterclockwise rotation about the origin, $(x,y)\rightarrow (-x,-y)$ when we rotate figure pqrs counterclockwise through an angle of 180° about the origin, the coordinates will become $p1( - 1, - 2),q1( - 2, - 1),r1( - 3, - 2),s1( - 3, - 3)$ next, we reflect in the y-axis by negating the x-coordinates of the resulting figure to obtain, $p'(1, - 2),q'(2, - 1),r'(3, - 2),s'(3, - 3)$ therefore the correct answer is c

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