NEED HELP QUICKLY!!

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

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

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) 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 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago

Increase $32 in the ratio 2:5 decrease 54 litter in the ratio 5:1

Novay_Z [31]

Answer:

Step-by-step explanation:

32 = 2 : 5

32/7 = 4

2 x 4 : 4 x 5

8 : 20

8 0

2 years ago

Algebra help? 343 = (x/6) ^2/3

Vladimir79 [104]

I can only assume that you meant, "Solve for x:"

Apply the exponent 3/2 to both sides of this equation. The result will be

3/2

343 = x/6.

Multiplying both sides by 6 isolates x:

3/2

6*343 = x Since 7^3 = 343, the expression for x

can be rewritten as

3/2

6*(7^3) = x which can be further simplified, as follows:

x = 6^(3/2)*7^(9/2), or:

x = 6^(3/2)*7^(8/2)*√7, or

x = 6^(3/2)*7^4*√7

3 0

3 years ago

What the percent change from 50; 35

il63 [147K]

50 -> 35=change by 15
change by 5=10%
so multiply by 3
3(5)=3(10%)
15=30%

5 0

3 years ago

a certain state the license plate has two digits, then two letters, and then two digits. How many different license plates are p

xxTIMURxx [149]

Step-by-step explanation:

Choosing 4 digits out of 10 = 10C4 = 210.

Arranging the 4 digits in 4 places = 4!

Total number of ways to arrange digits

= 210 * 4! = 5040.

Choosing 2 letters out of 26 = 26C2 = 325

Arranging the 2 letters in 2 places = 2!

Total number of ways to arrange letters

= 325 * 2! = 650.

Total number of possible license plates

= 5040 * 650 = 3,276,000.

7 0

3 years ago

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