In 42 minutes, Mr. Goo can run 6 miles. How long does it take him to run 1 mile?

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago

Help please !!!!!!!!!!!!!!

yKpoI14uk [10]

1) The triangles are similar, so correponding sides have the same ratio.
.. 3x/20 = (4x +2)/28
.. 3x/5 = (4x +2)/7 . . . . . . . multiply by 4
.. 7*3x = 5(4x +2) . . . . . . . multiply by 35 = 5*7
.. 21x = 20x +10 . . . . . . . . eliminate parentheses
.. x = 10 . . . . . . . . . . . . . . . subtract 20x

2) The triangles are similar, so corresponding measures have the same ratio.
.. x/42 = 10/15
.. x = 42*(10/15) = 28

3) The third angle of triangle ABC is 180° -20° -80° = 80°. Then two of the angles of ABC match those of JKL. The 3rd answer selection is appropriate.

5 0

3 years ago

Solve for "x" : 2x + 3x + 4x =36

Anestetic [448]

Answer:

4

Step-by-step explanation:

2x + 3x + 4x = 36

Evaluate like terms.

9x = 36

Find x.

x = 36 ÷ 9

x = 4

7 0

3 years ago

Read 2 more answers

Work out the value of 5x-2y when x=-2 and y=-3

max2010maxim [7]

I guess this how it would go
(5)(−2)−(2)(−3)

So your answer should be −4

3 0

2 years ago

Gloria has red, green, and purple sweatshirts. she has black, white, and grey sweatpants. how many different outfits are possibl

OLga [1]

Nine outfits because you do 3 x 3

6 0

2 years ago