Help with these 2 please

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

Four students were scheduled to give book reports in 1 hour. After the first report, 2/3 hour remained. The next two reports too

Novosadov [1.4K]

Answer:1/4 hour remaining

Step-by-step explanation:

Presenter 1 - 1/3 hour

Presenter 2 - 1/6 hour

Presenter 3 - 1/4 hour

Convert to least common denominator of 12

4/12 + 2/12 + 3/12 = 9/12

12/12 - 9/12 = 3/12

Reduce 3/12 to get 1/4 hour remaining

3 0

3 years ago

The ratio of basketball players to baseball players at Cypress High School is 4 to 3. If Coach Nelson has 20 basketball players,

Yuri [45]

Answer:

35

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Americans receive an average of 20 Christmas cards each year. Suppose the number of Christmas cards is normally distributed with

soldier1979 [14.2K]

The distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.

<h3>Probability</h3>

a. Distribution

X ~ N (20 , 6)

b. P(x ≤24)

= P[(x - μ ) / σ (24 - 20) / 6]

= P(z ≤0.67)

= 0.74857

=0.7486

Hence:

Probability = 0.7486

c. P(21 < x < 26)

= P[(21 - 26)/ 6) < (x - μ ) / σ < (24 - 20) / 6) ]

= P(-0.83 < z < 0.67)

= P(z < 0.67) - P(z < -0.)

= 0.74857- 0.2033

= 0.54527

Hence:

Probability =0.54527

d. Using standard normal table ,

P(Z < z) = 66%

P(Z < 0.50) = 0.66

z = 0.50

Using z-score formula,

x = z× σ + μ

x = 0.50 × 6 + 20 = 23

23 Christmas cards

Therefore the distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.

Learn more about probability here:brainly.com/question/24756209

#SPJ1

8 0

2 years ago

Estimate the difference of the decimals below by rounding to the nearest whole number. 23.827

iragen [17]

Step-by-step explanation:

The key rule is: to round up the number one digit to the left has to be the number 5 or more, else it rounds down.

23.827 rounds up because the 8 is more than 5.

1.218 rounds down because 2 is less than 5.

24-1= 23

6 0

3 years ago