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Tpy6a [65]
3 years ago
5

Conduct a test at the a=0.05 level of significance by determining (a) null and alternative hypothesis, (b) the test statistic, (

c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1 > p2. The sample data are x1=117 n1=249 x2=141 n2=312
Mathematics
1 answer:
raketka [301]3 years ago
8 0

Answer:

Since p > alpha, we accept H0, there is no evidence to prove that p1 is greater than p2

Step-by-step explanation:

Set up hypotheses as

H_0: p1 = p2\\H_a: p1>p2

(Right tailed t test)

Alpha = 0.05

Sample               I                II                Total

X                        117             141              248

N                        249           312            561

p                        0.4700     0.4519        0.4420

p difference = 0.0181    

Std dev = \sqrt{p(1-p)(\frac{1}{n_1} +\frac{1}{n_2} )}

=0.0427

z statistic = 0.424

p value = 0.33724

Since p > alpha, we accept H0, there is no evidence to prove that p1 is greater than p2

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Answer:

(a) Hence, the margin of error reported by The Marist Poll was correct.

(b) Based on a 95% confidence interval the poll does not provide convincing evidence that more than 70% of the population think that licensed drivers should be required to retake their road test once they turn 65.

Step-by-step explanation:

We are given that the Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age.

It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level.

(a) <u>Margin of error formula is given by;</u>

             Margin of Error =  Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }  

where, \alpha = level of significance = 1 - 0.95 = 0.05 or 5%

Standard of error =  \sqrt{\frac{\hat p(1-\hat p)}{n} }

Also, \hat p = sample proportion of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age = 66%

n = sample of American adults = 1.018

The critical value of z for level of significance of 2.5% is 1.96.

So, <em>Margin of Error </em>=  Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }  

                                =  1.96  \times \sqrt{\frac{0.66(1-0.66)}{1,018} } = 0.03 or 3%

Hence, the margin of error reported by The Marist Poll was correct.

(b) Now, the pivotal quantity for 95% confidence interval for the population proportion who think that licensed drivers should be required to retake their road test once they turn 65 is given by;

                   P.Q. =  \frac{\hat p-p}{ \sqrt{\frac{\hat p(1-\hat p)}{n} }}  ~ N(0,1)

So, <u>95% confidence interval for p</u> =  \hat p \pm \text{Margin of error}

                                                        =  0.66 \pm 0.03

                                                        =  [0.66 - 0.03 , 0.66 + 0.03]

                                                        =  [0.63 , 0.69]

Hence, based on a 95% confidence interval the poll does not provide convincing evidence that more than 70% of the population think that licensed drivers should be required to retake their road test once they turn 65 because the interval does not include the value of 70% or more.

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Answer:

William will make 12 bags of food and each of the bag will contains 2 cans of fruit and 5 cans of vegetables.

Step-by-step explanation:

Given:

Number of fruits cans = 24

Number of veggies cans = 60

William will have to distribute them in equal bags with equal cans of fruits and vegetables respectively.

For this:

We have to find the GCF (greatest common factor) of 24 and 60.

GCF by listing out the factors method.

Factors of 24  : 1,2,3,4,6,8,12,24

Factors of 60 : 1,2,3,4,5,6,10,12,15,20,30,60

So,

The greatest common factor of 24 and 60 is 12.

The number of bags William will used for equal distribution = 12

Now,

We have to distribute the veggies and fruits in equal number of cans to these 12bags.

Number of fruits cans used in each bag = \frac{24}{12} = 2

Number of vegetables can used in each bag = \frac{60}{12} =5

We can say that:

William will make 12 bags of food and each of the bag will contains 2 cans of fruit and 5 cans of vegetables.

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