Question

In 1945. an organization surveyed 1100 adults and asked. "Are you a total abstainer from, or do you on occasion consume, alcoholic beverages?" Of the 1100 adults surveyed. 363 indicated that they were total abstainers In a recent survey, the same question was asked of 1100 adults and 341 indicated that they were total abstainers Complete parts (a) and (b) below. The proportions of the adults who took the 1945 survey and the recent survey who were total abstainers are .33 and .31. respectively. (Round to three decimal places as needed) Has the proportion of adults who totally abstain from alcohol changed? Use the <<= 0.10 level of significance First verify the model requirements Select all that apply. The sample size is less than 5% of the population size for each sample The data come from a population that is normally distributed. The samples are independent The samples are dependent n_1p_1 (1 - p_1) greaterthanorequalto 10 and n_2p_2(1 - P_2) greaterthanorequalto 10 The sample size is more than 5% of the population size for each sample. Identify the null and alternative hypotheses for this test. Let p_1 represent the population proportion of 1945 adults who were total abstainers and p_2 represent the population proportion of recent adults who were total abstainers Determine the null and alternative hypotheses. H_0: P_1 = P_2 H_1: p_1 notequalto P_2 Find the test statistic for this hypothesis test. (Round to two decimal places as needed) Determine the P-value for this hypothesis test. (Round to three decimal places as needed) Interpret the P-value If the population proportions are equal. one would expect a sample difference proportion greater than the absolute value of the one observed in about out of 100 repetitions of this experiment. (Round to the nearest integer as needed.)

Answer 1

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.