How many numbers have an absolute value of 6

The estimate of the population proportion should be within plus or minus 0.02, with a 90% level of confidence. The best estimate

Rudiy27

Answer:

The sample size required is 910.

Step-by-step explanation:

The confidence interval for population proportion is:

$CI=\hat p\pm z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

The margin of error is:

$MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

Given:

$\hat p = 0.16\\MOE= 0.02\\Confidence\ level =0.90$

The critical value of z for 90% confidence level is:

$z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645$ *Use a standard normal table.

Compute the sample size required as follows:

$MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.02=1.645\times \sqrt{\frac{0.16(1-0.16)}{n} }\\n=\frac{(1.645)^{2}\times 0.16\times (1-0.16)}{(0.02)^{2}} \\=909.2244\\\approx910$

Thus, the sample size required is 910.

5 0

4 years ago

A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after

lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% 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 persons who were still employed primarily as engineers = $\frac{111}{200}$ = 0.555

n = sample of graduates = 200

p = population proportion of engineers

Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ , $0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ ]

= [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0

4 years ago

What is the value of n in the equation below?

Luden [163]

Answer:

I believe it is 12.

Step-by-step explanation:

12 - 3 = 9

7 0

3 years ago

(2,5) and (3,7)

zzz [600]

Answer:

equation in slope intercept form: y = 2x +1

Step-by-step explanation:

given coordinates = (2,5) and (3,7)

find slope,m = (y2 - y1) / (x2 - x1)

= (7 - 5) / ( 3 - 2)

= 2

using the formula: y - y1= m( x - x1 )
: y - 5 = 2( x - 2 )

: y = 2x +1

8 0

3 years ago

Read 2 more answers

What is a real world problem for 2(n+20)=110

sashaice [31]

John

You went to the store to buy a gift for your friend. Jeans are 20 dollars (what a deal) so you buy one for yourself and your friend. Then you see this really cute top and you want to match with your friend, so you get two of those too.

The total on the bill is $110.

While walking to your car, you lose the receipt.

When you give your friend the gift, she is excited. But says you spent too much money on her! She wants to pay you back for the shirt, but you can't remember how much it cost

6 0

4 years ago