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Reil [10]
3 years ago
7

The General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. If we wanted

to limit the margin of error of a 95% con dence interval to 2%, about how many Americans would we need to survey?
Mathematics
1 answer:
bazaltina [42]3 years ago
3 0

Answer: we need to survey 2285 Americans.

Step-by-step explanation:

When population proportion (p) is known, then the formula to evaluate the sample size( minimum) is given by :-

n=p(1-p)(\frac{z^*}{E})^2, where E = margin of error , z* = critical z-value.

Given: p= 0.61, E=0.02, Critical z value for 95% = 1.96

Required sample size: n=(\frac{1.96}{0.02})^2(0.61)(1-0.61)

=(98)^2(0.61)(0.39)\\\\=9604(0.61)(0.39)\\\\\approx$$2285

Hence, we need to survey 2285 Americans.

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Answer:

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So with 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% of significance the mean of the group A is significantly lower than the mean for the group B.  

Step-by-step explanation:

The statistic to test the hypothesis is given by this formula:

t=\frac{(\bar X_A-\bar X_B)-(\mu_{A}-\mu_B)}{\sqrt{\frac{s_A^2}{n_A}}+\frac{s_B^2}{n_B}}

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The system of hypothesis on this case are:

Null hypothesis: \mu_A \leq \mu_B

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Null hypothesis: \mu_A - \mu_B \leq 0

Alternative hypothesis: \mu_A -\mu_B>0

Our notation on this case :

n_A =50 represent the sample size for group A

n_B =40 represent the sample size for group B

\bar X_A =280 represent the sample mean for the group A

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t=\frac{(280 -250)-(0)}{\sqrt{\frac{20^2}{50}}+\frac{23^2}{40}}=6.511

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p_v =P(t_{88}>6.960) =2.91x10^{-10}

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