Round to the place value of the underlined digit 301.587

In 1945. an organization surveyed 1100 adults and asked. "Are you a total abstainer from, or do you on occasion consume, alcohol

oee [108]

Answer:

$n_1 p_1 =1100*0.31=341 \geq 10$

$n_1 (1- p_1) =1100*(1-0.31)=759 \geq 10$

$n_2 p_2 =1100*0.33=363 \geq 10$

$n_2(1- p_2) =1100*(1-0.33)=737 \geq 10$

The data come from a population that is normally distributed.

The samples are independent

$z=\frac{0.31-0.33}{\sqrt{0.32(1-0.32)(\frac{1}{1100}+\frac{1}{1100})}}=-0.47$

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

$p_v =2*P(Z$

If the population proportions are equal one would expect a sample difference proportion greater than the absolute value of the 64 observed in about out of 100 repetitions of this experiment.

Step-by-step explanation:

1) Data given and notation

$X_{1}=341$ represent the number of people indicated that they were total abstainers In a recent survey

$X_{2}=363$ represent the number of people indicated that they were total abstainers In a 1945 survey,

$n_{1}=1100$ sample 1

$n_{2}=1100$ sample2

$p_{1}=\frac{341}{1100}=0.31$ represent the proportion of people indicated that they were total abstainers In a recent survey

$p_{2}=\frac{363}{1100}=0.33$ represent the proportion of people indicated that they were total abstainers In a 1945 survey,

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

$n_1 p_1 =1100*0.31=341 \geq 10$

$n_1 (1- p_1) =1100*(1-0.31)=759 \geq 10$

$n_2 p_2 =1100*0.33=363 \geq 10$

$n_2(1- p_2) =1100*(1-0.33)=737 \geq 10$

The data come from a population that is normally distributed.

The samples are independent

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference in the two proportions of interest, 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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{361+343}1100+1100}=0.32$

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:

$z=\frac{0.31-0.33}{\sqrt{0.32(1-0.32)(\frac{1}{1100}+\frac{1}{1100})}}=-0.469$

Statistical decision

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

$p_v =2*P(Z$

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that we have a significant differences between the two proportions.

If the population proportions are equal one would expect a sample difference proportion greater than the absolute value of the 64 observed in about out of 100 repetitions of this experiment.

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Anna [14]

Answer:

B

Step-by-step explanation:

Got it right on edge

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