Recall the real number line

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! Algebraic Relationship y= x/4-1

loris [4]

32, 7

0, - 1

20, 4

You just plug in the numbers and solve

3 0

3 years ago

⚠️URGENT PLS HELPPPPP LINKS WILL BE REPORTED

balu736 [363]

The answer is the last one

6 0

3 years ago

If angle 1 is 140°, then find the measure of the other angles.

Vlad [161]

<h2>Angles</h2>

<h3>If angle 1 is 140°, then find the measure of the other angles.</h3>

∠2 = 40°
∠3 = 40°
∠4 = 140°
∠5 = 140°
∠6 = 40°
∠7 = 40°
∠8 = 140°

Explanation:

The relationship between ∠1 and ∠2 are supplementary angles, so when you add up their measurements, it will become 180°. Simply subtract 180 and 140 to get the measure of ∠2. As well as ∠3, they're linear pairs. And they are also supplementary. To determine the measure of ∠6 and ∠7, notice the relationship between ∠2 and ∠6. As you noticed, it is corresponding angles. So they have the same measurement. If ∠2 = 40°, then ∠6 = 40°. As well as ∠7, because the relationship between ∠6 and ∠7 are vertical pairs. So the angle measurement of ∠7 is also 40°.

Meanwhile, the relationship between ∠1 and ∠4 are vertical pairs. It means they also have the same measurement. So ∠4 = 140°. The relationship between ∠1 and ∠5 are corresponding angles, so they also have the same measurement. If ∠1 = 140°, then ∠5 = 140°. The relationship between ∠1 and ∠8 are alternate exterior angles, and they also have the same measurement. If ∠1 = 140°, then ∠8 = 140°.

Wxndy~~

8 0

2 years ago

On September 25, 2014 Colson Corp. sold 200,000 widgetrons to Cavanaugh Corp at $5 per unit. Half of the units were delivered on

borishaifa [10]

On September 25.2014 Colson Corp. sold 200.000 widgetrons for 5$ per unit.
200.000*5$=1.000.000$. Same day company was paid for 40% which is 400.000$.
The rest of the money he get paid in two equal parts on November 15.2014 and January 20.2015. So Calson Corp. was paid 300.000$ on November 15.2014 and 300.000$ on January 20.2015.

6 0

3 years ago