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Suppose we are testing people to see if the rate of use of seat belts has changed from a previous value of 88%. Suppose that in

Andreas93 [3]

Answer:

a) We would expect to see 500*0.88=440

b) $z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376$

$p_v =2*P(Z>1.376)=0.167$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when $p_v$ we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=450 represent the people that have the seat belt fastened

$\hat p=\frac{450}{500}=0.9$ estimated proportion of people that have the seat belt fastened

$p_o=0.88$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) Part aWe would expect to see 500*0.88=440Part bConcepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion changes fro m 0.88.: Null hypothesis:[tex]p=0.88$

Alternative hypothesis: $p \neq 0.88$

When we conduct a proportion test we need to use the z statisitc, 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 requires we can replace in formula (1) like this:

$z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376$

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 assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z>1.376)=0.167$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when $p_v$ we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".

8 0

3 years ago

1 point

Margarita [4]

Answer:

35% or 0.35

Step-by-step explanation

If 20 grams is 100%, that means each gram is 5% of the whole. So 7 time 5 is 35.

4 0

4 years ago

Please enter the missing number: 4, 8, 14, 22, ?

lara31 [8.8K]