Nikhil drew a picture of an open greeting card on his paper. What is the total area of the drawing of the open greeting card?

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

Yall im almost done pls HELP

S_A_V [24]

Answer:

61 degrees

Step-by-step explanation:

==>Given ∆MNO,

MO = 18,

MN = 6

m<O = 17°

==>Required:

Measure of <N

==>SOLUTION:

Use the sine formula for finding measure of angles which is given as: Sine A/a = Sine B/b = Sine C/c

Where,

Sine A = 17°

a = 6

Sine B = N

b = 18

Thus,

sin(17)/6 = sin(N)/18

Cross multiply

sin(17)*18 = sin(N)*6

0.2924*18 = 6*sin(N)

5.2632 = 6*sin(N)

Divide both sides by 6

5.2632/6 = sin(N)

0.8772 = sin(N)

sin(N) = 0.8772

N = sin^-1(0.8772)

N ≈ 61° (approximated)

8 0

3 years ago

a Baker's going to make 24 blueberry pie she wants to make sure that each by contains 3.5 cups of blueberries how many cups of b

Leokris [45]

She is going to need 84 cups of blueberries. 3.5*24= 84 i believe :)

8 0

3 years ago

Mai had $14.50. She spent $4.35 at the snack bar and $5.25 at the arcade. What is the exact amount of money mai has left

velikii [3]

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ Mai has $4.90 left.

Add $4.35 and $5.25 to get the total amount that she spent:

$4.35 + $5.25 = $9.60

Then subtract to get the amount of money left:

$14.50 - $9.60 = $4.90

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

3 0

3 years ago

Read 2 more answers

A cafeteria manager can choose from among six side dishes for the lunch menu: applesauce, broccoli, corn, dumplings, egg rolls,

pochemuha

Answer:

20%

brainly.com/question/12746119

5 0

2 years ago

Read 2 more answers