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Zanzabum
3 years ago
11

An appropriate metric unit for the mass of an eyelash.

Mathematics
1 answer:
siniylev [52]3 years ago
7 0

Answer: Mass of a typical house cat:

Automatically eliminate B and D since they are not metric unit for mass

So, that leaves us with A and C. And since a cat is not huge nor weighs that much, the most reasonable one would be to measure it in grams.

Correct answer: C

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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".  

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