PLEASE HELP!! thank you!!!

Need help on this . Explain how you got it ?

kvasek [131]

The goal is to find a number between 0.125 and 0.18
1/5 is 0.2
0.2, 1.6, 0.09 are not in the range. Therefore, the square root of 0.02 (0.1414) is the answer.

8 0

3 years ago

Consider this equation: 2x + 2 = 11 - x

77julia77 [94]

Addition bc. you need adding to both sides x and will get this result of 3x+2 = 11

hope helped

4 0

3 years ago

Read 2 more answers

spin [16.1K]

Answer:

$z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67$

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

$\hat p=\frac{18}{100}=0.18$ estimated proportion of ounce cups of coffee that were underfilled

$p_o=0.1$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion it's higher than 0.1 or 10%:

Null hypothesis: $p\leq 0.1$

Alternative hypothesis: $p > 0.1$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67$

4) 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 one right tailed test the p value would be:

$p_v =P(Z>2.67)=0.0037$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ 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 thetrue proportion is not significanlty higher than 0.1 or 10% .