Giving brainlist please help me out

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

Simplify the expression given below. (732 - 83 + 21) + (-1363 - 11.2 + 133 + 19)

Vera_Pavlovna [14]

Answer:

732−83+21−1363−11.2+133+19= −552.2

8 0

2 years ago

Read 2 more answers

What is the area of this triangle x1,y1)(x3,y3)1,y1)

Novosadov [1.4K]

Please give it's proper dimensions..

Thanks!

7 0

2 years ago

Isabella has a board that is 5.24 feet long and 1.34 feet wide.

Masteriza [31]

Answer:

Length of 1 piece= Length of total divided by 8

Step-by-step explanation:

Length of total= 5.24 feet long

Width of total= 1.34 feet wide

Width is divided by 1 line so, there are 2 parts.

Width= 1.34 divided by 2

Width= 0.67 feet wide

Length is divided by 7 lines, so there are 8 parts.

Length= 5.24 divided by 8

Length=0.655 feet long

They are asking for formula to find the length of each piece

Length of 1 piece= Length of total divided by 8

Hope that helps!

8 0

2 years ago

The value of which of these expressions is closest to e?

mars1129 [50]

Answer:

B.

Step-by-step explanation:

Given: expressions

To find: the expression whose value is closest to e

Solution:

Value of e is 2.718

$\left ( 1+\frac{1}{19} \right )^{19}=\left ( \frac{20}{19} \right )^{19}=2.65\\\left ( 1+\frac{1}{22} \right )^{22}=\left ( \frac{23}{22} \right )^{22}=2.659\\\left ( 1+\frac{1}{20} \right )^{20}=\left ( \frac{21}{20} \right )^{20}=2.653\\\left ( 1+\frac{1}{21} \right )^{21}=\left ( \frac{22}{21} \right )^{21}=2.656$

Therefore, value 2.659 is closest to the value of e.

So, option B. is correct.

7 0

3 years ago