Mathematics, 17.03.2020 00:00 elizabethhubbe

In a survey of 450 adults 18 to 29 years of age, 419 said they use the Internet. In a survey of 400 adults 30 to 49 years of age, 324 said they use the internet. At a=0.01, can you reject the claim that the proportion of internet users is the same for two age groups?

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Answer from: Apple557

$z=\frac{0.931-0.81}{\sqrt{0.874(1-0.874)(\frac{1}{450}+\frac{1}{400})}}=5.306$

$p_v =2*P(Z5.306) =1.12x10^{-7}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the two proportions are different at 1% of significance

Step-by-step explanation:

Data given and notation

$X_{1}=419$ represent the number of adults that use internet between 18-29 years

$X_{2}=324$ represent the number of adults that use internet between 30-49 years

$n_{1}=450$ sample selected

$n_{2}=400$ sample selected

$\hat p_{1}=\frac{419}{450}=0.931$ represent the proportion of adults that use internet between 18-29 years

$p_{2}=\frac{324}{400}=0.81$ represent the proportion of adults that use internet between 30-49 years

z would represent the statistic (variable of interest)

p_v represent the value for the test (variable of interest)

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} - p_{2}=0$

Alternative hypothesis: $p_{1}- p_{2} \neq 0$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{419+324}{450+400}=0.874$

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.931-0.81}{\sqrt{0.874(1-0.874)(\frac{1}{450}+\frac{1}{400})}}=5.306$

Statistical decision

Since is a two sided test the p value would be:

$p_v =2*P(Z5.306) =1.12x10^{-7}$

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in fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

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In a survey of 450 adults 18 to 29 years of age, 419 said they use the Internet. In a survey of 400...

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