Mathematics, 07.04.2020 16:14 michellemonchez103

A cellphone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 135 of the subscribers would upgrade to a new cellphone at a reduced cost. At the 0.05 level of significance, is there evidence that more than 20% of the customers would upgrade to a new cellphone at a reduced cost

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Answer from: applesass

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

p_v =P(z3.913)=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(z3.913)=0.000046

Answer from: Quest

step-by-step explanation:

options

Answer from: Quest

the total estimated nappy use.=10,500,000

step-by-step explanation:

given that on an average the average baby uses 3528 napkins.

in a particular year 2014, the number of births were 299697 in new south wales.

rounding off the number of births to the nearest hundred thousand we have

no of birtsh = 300000

also rounding off the average usage to nearst thousand we get

average use of a baby = 3500

hence total number of nappy estimated to be used = 3500x300000

= 10,500,000

answer is 10,500,000 nappies

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