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Vesna [10]
2 years ago
13

What is a mixed number between 3 and 4

Mathematics
1 answer:
bulgar [2K]2 years ago
3 0

Answer:

ksdudjfudisotowts484slst

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bh

6, 17

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Step-by-step explanation:

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3 years ago
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In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a samp
dsp73

Answer:

(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401  

(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380  

We are confident at 95% that the difference between the two proportions is between -0.4401 \leq p_B -p_A \leq -0.1380

1.  -.4401 ≤ p1 - p2 ≤ -.1380

4.  The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

Step-by-step explanation:

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.

1.  -.4401 ≤ p1 - p2 ≤ -.1380

2.  -.4401 ≤ p1 - p2 ≤ .1380

3.  The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%

4.  The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

5.  The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Solution to the problem

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_1 represent the real population proportion for San Jose

\hat p_1 =\frac{30}{73}=0.411 represent the estimated proportion for San Jos

n_1=73 is the sample size required for San Jose

p_2 represent the real population proportion for San Francisco

\hat p_2 =\frac{56}{80}=0.7 represent the estimated proportion for San Francisco

n_2=80 is the sample size required for San Francisco

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_1 -\hat p_1) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401  

(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380  

We are confident at 95% that the difference between the two proportions is between -0.4401 \leq p_B -p_A \leq -0.1380

Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.

Based on this the correct options are:

1.  -.4401 ≤ p1 - p2 ≤ -.1380

4.  The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

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bonufazy [111]

The answer is A

1 1/2 *2 *1 1/2 = 4.5 or 4 1/2

5 0
2 years ago
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HELP!!! STATS!!!
Vadim26 [7]

Answer:

The 99% confidence interval is   165.776  <  \mu <  180.224  

Step-by-step explanation:

From the question we are told that

     The sample size is  n  = 100

     The sample mean is  \=  x  =  173 \  days

      The population standard deviation is  \sigma =  28 \ days  

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

      E = 2.58  *  \frac{28 }{\sqrt{100} }

=>    E = 7.224

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>     173 -7.224  <  \mu < 173 + 7.224    

=>     165.776  <  \mu < 180.224  

     

4 0
2 years ago
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