Which expression represents the algebraic phrase “nine times a number plus thirteen”?

It would be C)15.75 miles because you divide it to get 2 hrs and divide again to get an hour then again to get half and hour and you add on the last number and 2nd to last so.. 42/2 is 21/2 is 10.5/2 is 5.25 add 10.5 and 5.25 is 15.75 which is 15 and 3/4 miles :)).

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3 years ago

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A cellphone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new

stiv31 [10]

Answer:

$z=\frac{0.27 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=3.913$

$p_v =P(z>3.913)=0.000046$

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 the proportion of subscribers that would upgrade to a new cellphone at a reduced cost is significantly higher than 0.2 or 20%

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=135 represent the subscribers that would upgrade to a new cellphone at a reduced cost

$\hat p=\frac{135}{500}=0.27$ estimated proportion of subscribers that would upgrade to a new cellphone at a reduced cost

$p_o=0.2$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.2 or not.:

Null hypothesis: $p \leq 0.2$

Alternative hypothesis: $p > 0.2$

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 required we can replace in formula (1) like this:

$z=\frac{0.27 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=3.913$

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 provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>3.913)=0.000046$

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 the proportion of subscribers that would upgrade to a new cellphone at a reduced cost is significantly higher than 0.2 or 20%

6 0

4 years ago