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2 years ago

5

There are approximately 1.06 quarts per liter. There are 4 quarts per gallon. If gasoline cost $0.50 per liter, how much does ga

soline cost per gallon?
A. $1.42/gal

B. $1.75/gal

C. $1. 89/gal

D. $2. 12/gal

E. $4.24/gal

2 answers:

mel-nik [20]2 years ago

7 0

Answer:

Im pretty sure its E

Step-by-step explanation:

hope this is right

Tanya [424]2 years ago

3 0

Answer:

E......?...........................

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$p_v =P(Z$

c) If we compare the p value obtained 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% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

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Step-by-step explanation:

Data given and notation

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

$\hat p=0.35$ estimated proportion of business owners plan to provide a holiday gift to their employees

$p_o=0.46$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:

Null hypothesis: $p\geq 0.46$

Alternative hypothesis: $p < 0.46$

When we conduct a proportion test we need to use the z statisitc, 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:

$z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710$

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.

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