Mathematics, 29.10.2020 16:30 jaz4499

Suppose that a sample of n students at the same university (instead of n) determines that % of the sample use the web browser. At the level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of %?

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Answer from: sophiee1021

Complete Question

The worldwide market share for a web browser was 20.5% in a recent month. Suppose that a sample of 200 random students at a certain university finds that 50 use the browser.

At the 0.05 level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.5%?

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the market share for the web browser at the university is greater than the worldwide market share of 20.5 %?

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.205

The sample size is n = 200

The number that uses the browser is k = 50

The level of significance is $\alpha = 0.05$

Generally the sample proportion is mathematically represented as

$\^ p = \frac{50}{200}$

=> $\^ p = 0.250$

The null hypothesis is H_o : p = 0.205

The alternative hypothesis is H_a : p 0.205

Generally the test statistics is mathematically represented as

$z = \frac{ \^ p - p }{ \sqrt{ \frac{ p (1 - p )}{ n} } }$

=> $z = \frac{ 0.250 - 0.205 }{ \sqrt{ \frac{ 0.205 (1 - 0.205 )}{ 200} } }$

=> z = 1.5764

Generally the area under z = 1.5764 to the right is

p-value = P( Z 1.5764 ) = 0.057467

From the value we obtained we can see that the $p-value \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the market share for the web browser at the university is greater than the worldwide market share of 20.5 %?

Answer from: Quest

The first digit is in the middle school 100s place

Answer from: Quest

see attached

step-by-step explanation:

given the function as;

$f(x)=2x^3-7x^2+2x+3$

write the equation replacing f(x) with y

$y=2x^3-7x^2+2x+3$

form and fill a table containing values of y with corresponding values of x

you can take a range for values of x as -3≤x≤3

solving for x=-3

$f(-3)=2*-3^3-7*-3^2+2*-3+3=2*-27-63+-6+3=-54-63-6+3y=-120$

for x=-2

$f(-2)=2*-2^3-7*-2^2+2*-2+3=2*-8-28+-4+3=-16-28-4+3=-45$

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values of x and obtain the following graph

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