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CaHeK987 [17]
3 years ago
7

PLEASE HELP WILL GIVE BRAINLIEST!!!!!

Mathematics
2 answers:
Zolol [24]3 years ago
8 0

B is the correct answer. I hope this helps!

lutik1710 [3]3 years ago
7 0

B is right and give jilko505 brainliest.

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Suppose you are working in an insurance company as a statistician. Your manager asked you to check police records of car acciden
pochemuha

Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

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

where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

           p = population percentage of all car accidents

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }} < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}]

  = [ 0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} , 0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} ]

  = [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

<u><em>Let p = percentage of teenagers who where involved in car accidents</em></u>

So, Null Hypothesis, H_0 : p = 18%    {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, H_A : p \neq 18%    {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is <u>One-sample z proportion statistics</u>;

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

where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

So, <u><em>test statistics</em></u>  =  \frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}  

                              =  1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.

4 0
3 years ago
Joel looks at a map of Colorado, scowls, and tears it to shreds. "I don't need this lousy map," he mutters, puffing out his ches
Andru [333]

Answer:12 x 9

Step-by-step explanation: 600 x 2/100 = 1200/100 = 12

450 x 2/100 = 900/100 = 9

Remember that 100km = 2 units

5 0
3 years ago
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Ron needed to buy a new jacket for his upcoming ski trip. The ticket price of the jacket was $199.00. After the sales tax was ad
Levart [38]

Answer: 5%

Step-by-step explanation:

Given that:

Ticket price of jacket = $199

Final price after addition of sales tax = $208.95

Percentage sales tax on the jacket =?

Final price after addition of sales tax = price of jacket + amount of sales tax paid

208.95 = 199 + amount of sales tax paid

amount of sales tax paid = 208.95 - 199

amount of sales tax paid = $9.95

Let Percentage of sales tax = x

(x / 100) * 199 = 9.95

199x / 100 = 9.95

199x = 995

x = 5

Hence percent sales tax paid = 5%

8 0
3 years ago
Factor completely or identify as prime.
SSSSS [86.1K]

Answer:

Factor the polynomial

C. (p-3) (4p+5) (4p-5)


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3 years ago
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Schach [20]

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The 3 lb bag of cat food would last for 4 weeks.

Step-by-step explanation:

6 0
3 years ago
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