Somebody Please Help, I'm Super Confused

A large manufacturer that sells consumer products on-line wishes to publicize its customer satisfaction in an advertisement. Spe

sergejj [24]

Answer:

$z=\frac{0.93 -0.9}{\sqrt{\frac{0.9(1-0.9)}{400}}}=2$

$p_v =P(z>2)=0.0228$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who say yes is significantly higher than 0.9 or 90%.

Step-by-step explanation:

Data given and notation

n=400 represent the random sample taken

X=372 represent the number of people who say yes

$\hat p=\frac{372}{400}=0.93$ estimated proportion of people who say yes

$p_o=0.9$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that that more than 90% of its costumers would tell their friend to buy a product from the manufacturer.:

Null hypothesis: $p\leq 0.9$

Alternative hypothesis: $p > 0.9$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.93 -0.9}{\sqrt{\frac{0.9(1-0.9)}{400}}}=2$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed for this case is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2)=0.0228$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who say yes is significantly higher than 0.9 or 90%.

8 0

3 years ago

Use the quadratic function f(x) = 2x^2 + 4x - 15. Find any x intercepts.

8_murik_8 [283]

So for this function we will be using the quadratic formula, which is $x=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a}$ , to solve. a = x^2 coefficient, b = x coefficient, and c = constant. Using our equation, we can solve for the zeros (x-intercepts) as such:

$x=\frac{-4+\sqrt{4^2-4*2*(-15)}}{2*2},\frac{-4-\sqrt{4^2-4*2*(-15)}}{2*2}\\ \\ x=\frac{-4+\sqrt{16-(-120)}}{4},\frac{-4-\sqrt{16-(-120)}}{4}\\ \\ x=\frac{-4+\sqrt{136}}{4},\frac{-4-\sqrt{136}}{4}\\ \\ x=1.92,-3.92$