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Oksana_A [137]
3 years ago
5

drug sniffing dogs must be 95% accurate in their responses because their handlers don't want them to miss durgs and also don't w

ant false positives. a new dog is being tested and is right in 46 of 50 trials. find the 95% confidence interval for the proportion of times the dog will be correct
Mathematics
1 answer:
GenaCL600 [577]3 years ago
7 0

Answer:

95% Confidence interval:  (0.8449,0.9951)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 50

Number of times the dog is right, x = 46

\hat{p} = \dfrac{x}{n} = \dfrac{46}{50} = 0.92

95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

Putting the values, we get:

0.92 \pm 1.96(\sqrt{\dfrac{0.92(1-0.92)}{50}})\\\\ = 0.92\pm 0.0751\\\\=(0.8449,0.9951)

(0.8449,0.9951) is the required 95% confidence interval for the proportion of times the dog will be correct.

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An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t
erastova [34]

Answer:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

\bar X=46.5 represent the mean

\sigma=1.1 represent the population standard deviation

n=150 sample size  

\mu_o =46.7 represent the value to verify

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:  

Null hypothesis:\mu = 46.7  

Alternative hypothesis:\mu \neq 46.7  

Since we know the population deviation the statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing we got:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

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Answer:

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