I need help to find the equation and measure pls.

A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a populat

eimsori [14]

Answer:

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

$p_v =P(t_{15}>2)=0.0320$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.

Step-by-step explanation:

1) Data given and notation

$\bar X=25$ represent the sample mean

$s=2$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =24$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Confidence =0.95 or 95%

$\alpha=0.05$

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:

Null hypothesis: $\mu \leq 24$

Alternative hypothesis: $\mu > 24$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>2)=0.0320$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.

3 0

3 years ago

What is 16500 rounded to the nearest ten thousand?

frosja888 [35]

16500 Rounded to the nearest ten thousand is 20,000

4 0

3 years ago

What number does x stand for in this equation?

Alex777 [14]

Answer:

C. -3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

-7x + 6 = 27

Step 2: Solve for x

[Subtraction Property of Equality] Subtract 6 on both sides: -7x = 21
[Division Property of Equality] Divide -7 on both sides: x = -3

3 0

2 years ago

Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on

boyakko [2]

Answer:

Mathew invested $600 and $2400 in each account.

Solution:

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest $(I_1)$ earned in first account for one year,

$\text {simple interest}=\frac{\text {pnr}}{100}$

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get $(I_1)$ as,

$I_{1}=\frac{P \times 1 \times 3}{100}$ ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest $(I_2)$ earned in second account,

$I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2$

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

$I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3$

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

$I = I_1+ I_2 ---- eqn 4$

By substituting eqn 1 , 2, 3 in eqn 4

$\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}$

$\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}$

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600

Hence, Mathew invested $600 and $2400 in each account.

3 0

3 years ago

Read 2 more answers

Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0 = 2x2 + 3x – 8? Quadratic formula: x

Lynna [10]

ANSWER

1.39

EXPLANATION

The given quadratic equation is

$0 = 2 {x}^{2} + 3x - 8$

This is the same as,

$2 {x}^{2} + 3x - 8 = 0$

Comparing to

$a {x}^{2} + bx + c = 0$

We have

a=2, b=3,c=-8

Using the quadratic formula, the solution is given by:

$x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

We substitute the values to get,

$x = \frac{ - 3\pm \: \sqrt{ {3}^{2} - 4(2)( - 8)} }{2(2)}$

$x = \frac{ - 3\pm \: \sqrt{ 73} }{4}$

The positive root is

$x = \frac{ - 3 + \: \sqrt{ 73} }{4} = 1.39$

to the nearest hundredth.

4 0

3 years ago

Read 2 more answers