What is the slope of the line that passes through (4, 12) and (5, 16)?

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

4 years ago

1 Two high school basketball teams have identical records of 15 wins and 2 losses. Sunnyside High School's mean score is 50 poin

yanalaym [24]

Answer:

Lin likes Sunnyside High School

Step-by-step explanation:

The Mean Absolute Deviation (MAD) is a measure of how dispersed the values are, in this case the score of the teams. So, Sunnyside High School, with a low MAD, scores nearly the same amount of points in every game.

6 0

4 years ago

How to find the volume of a rectangular prism

mr_godi [17]

Length times width times height

8 0

3 years ago

Read 2 more answers

Hello people ~<br>factorise : 5x²+16x+3

son4ous [18]

5x²+16x+3
5x²+15x+x+3
5x(x+3)+1(x+3)
(5x+1)(x+3)

Done!

5 0

2 years ago

Read 2 more answers

L.F. Richardson proposed a model to describe the spread of war fever. His proposal: If y(t) is the percent of the population adv

Lena [83]

Answer:

y'(t)=ky(t)(100-y(t))

Step-by-step explanation:

The rate of change of y(t) at any time is the derivative of y with respect to time y, y'(t)

If y(t) is the percent of the population advocating war at time t

then 100-y(t) is the percent of the population not advocating war

The product of the percentage of the population advocating war and the percentage not advocating war would be

y(t)(100-y(t))

If the rate of change of y(t) at any time is proportional to the product of the percentage of the population advocating war and the percentage not advocating war, then

y'(t)=ky(t)(100-y(t))

where <em>k is the constant of proportionality </em>

8 0

3 years ago