Evaluate 5x for each value of x x = 0.3

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

The slope of a line passing through the points (-2, 3) and (2, 7) is 1. a)True b) False

Paraphin [41]

<h3>Answer: True</h3>

Slope formula

m = (y2 - y1)/(x2 - x1)

m = (7 - 3)/(2 - (-2))

m = (7 - 3)/(2 + 2)

m = 4/4

m = 1

7 0

2 years ago

Read 2 more answers

Find the value of c and AB if B is between A and C,AB=2r,AC=-5and BC=7

elena-14-01-66 [18.8K]

AB=2r

BC=7

AC=AB+BC

-5=2r+7

-12=2r

-6=r

AB=-12

6 0

3 years ago

Use the distributive property to simplify 1/6 × (36+1/2)

kicyunya [14]

1/6(36 + 1/2) = 6 + 1/12 = 72/12 + 1/12 = 73/12

5 0

2 years ago

Express in the simplest form 777:147

Dvinal [7]

Answer:

your answer should be 37:7

Try to reduce the ratio further with the greatest common factor (GCF).

The GCF of 777 and 147 is 21

Divide both terms by the GCF, 21:

777 ÷ 21 = 37

147 ÷ 21 = 7

The ratio 777 : 147 can be reduced to lowest terms by dividing both terms by the GCF = 21 :

777 : 147 = 37 : 7

Therefore:

777 : 147 = 37 : 7

3 0

3 years ago

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