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3 years ago

5

You are given the difference of the numbers of boys and girls in a class and the ratio of boys to girls. How many boys and how m

any girls are in the class?
4 more girls; 9 for every 13

1 answer:

Tema [17]3 years ago

7 0

Answer:

Am sure boys are 13and girls also 13

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Step-by-step explanation:

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What is the HIGHEST common factor<br> of the set of numbers (54, 72, 90}?

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4 years ago

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

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

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

<u>95% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

6 0

3 years ago