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igomit [66]
3 years ago
14

How many odd integers are there between 101 and 199?​

Mathematics
2 answers:
Rudiy273 years ago
5 0

Answer:

98

Step-by-step explanation:

im guessing for points

Nonamiya [84]3 years ago
3 0

Answer:

49 odd integers

Step-by-step explanation:

since 199-101 is 98 so there is 98 integers between 101 and 199

then we have to divide 98 by 2 to find out how much odd(or even) integers are between 101 and 199

since 98 divided by 2 is 49 there are 49 odd integers between 101 and 199

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Answer:

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Step-by-step explanation:

Previous concepts and data given

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".  

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p_A represent the real population proportion of women age 18 to 30  agreed with the statement that a woman should have the right to a legal abortion for any reason

\hat p_A =\frac{26}{75}=0.347 represent the estimated proportion of women age 18 to 30  agreed with the statement that a woman should have the right to a legal abortion for any reason

n_A=75 is the sample size for A

p_B represent the real population proportion for women age 58 to 70  agreed with the statement that a woman should have the right to a legal abortion for any reason

\hat p_B =\frac{21}{64}=0.328 represent the estimated proportion of women age 58 to 70  agreed with the statement that a woman should have the right to a legal abortion for any reason

n_B=64 is the sample size required for B

z represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:  

Null hypothesis:p_{A} = p_{B}  

Alternative hypothesis:p_{A} \neq p_{B}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}   (1)

Where \hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236  

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Since is a two sided test the p value would be:  

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Part b  

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z_{\alpha/2}=2.58  

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(0.347-0.328) - 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=-0.188  

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And the 99% confidence interval for the difference of proportions would be given (-0.188;0.226).  

We are confident at 99% that the difference between the two proportions is between -0.188 \leq p_B -p_A \leq 0.226

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