"if a 40-pound weight stretches the spring 4 inches, ,find the distance that a 45-pound weight stretchs the spring"

Researchers conducted a study to find out if there is a difference in the use of ereaders by different age groups. randomly sele

ikadub [295]

No, there is significant difference in the use of e readers by different age groups.

Given sample 1 ( 29 years old) $n_{1}$ =628, $p_{1}$ =7%, sample 2( 30 years old) $n_{2}$ =2309, $p_{2}$ =0.11.

We have to first form hypothesis one null hypothesis and other alternate hypothesis.

$H_{0}:$ π1-π2=0

$H_{1}:$ π1-π2≠0

α=0.05

Difference between proportions $p_{1}-p_{2} =-0.04$

$p_{d}=0.07-0.11=-0.04$

The pooled proportion needed to calculate standard error is:

$p=(X_{1} -X_{2} )/(n_{1} +n_{2} )$

=(44+254)/(628+2309)

=0.10146

The estimated standard error of difference between means is computed using the formula:

$S_{p_{1} -p_{2} }=\sqrt{p(1-p)/m_{1}+p(1-p)/n_{2} }$

= $\sqrt{0.101*0.899/628+0.101*0.899/2309}$

= $\sqrt{0.000143+0.00003}$

= $\sqrt{0.000173}$

=0.01315

Z= Pd-(π1-π2)/ $S_{p_{1} -p_{2} }$

=-0.04-0/0.013

=-3.0769

This test is a two tailed test so the p value for this test is calculated as (using z table)

p value:2 P(Z<-3.0769)

=2*0.002092

=0.004189

P value< significance level of 5%.

Hence there is enough evidence to show the claim that there is a significant difference in the use of e readers by different age groups.

Learn more about hypothesis at brainly.com/question/11555274

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Question is incomplete as it also includes:

Significance level of 5%.

4 0

2 years ago

Ask Your Teacher Consider a binomial random variable with n = 5 and p = 0.8. Let x be the number of successes in the sample. Eva

Readme [11.4K]

Answer:

0.942 is the required probability.

Step-by-step explanation:

We are given the following in the question:

x is a binomial random variable with n = 5 and p = 0.8.

Then,

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

$P(x \geq 3) = P(x = 3) + P(x = 4) +P(x=5)\\\\= \binom{5}{3}(0.8)^3(1-0.8)^2 +\binom{5}{4}(0.8)^4(1-0.8)^1 +\binom{5}{5}(0.8)^5(1-0.8)^0\\\\= 0.2048 +0.4096+0.3276=0.942$

0.942 is the required probability.

4 0

3 years ago

What is the simplified expression for 3 a squared + 9 a b + 5 minus 4 a squared minus 4 a b + 3? Negative a squared + 5 a b + 8

MArishka [77]

Answer:

The answer is option 1.

Step-by-step explanation:

In order to simplify the expressions, you have to collect like terms :

3a² + 9ab + 5 - 4a² - 4ab + 3

= (3a² - 4a²) + (9ab - 4ab) + (5 + 3)

= -a² + 5ab + 8

7 0

2 years ago

Read 2 more answers

What is the width of a rectangle with length 5.5 cm and area 220 cm???

Aleks04 [339]

40 cm. Just divide the area by the length. Hope this helps!

7 0

3 years ago

Can the triangles be proven similar by AA?

vekshin1

Answer:

No, similar triangles cannot contain a pair of parallel lines. Yes, because || . Yes, because ∠QUR ≅ ∠TUS (vertical angles) and ∠R ≅ ∠S (alternate interior angles).

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers