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Ad libitum [116K]
3 years ago
15

What's is the numerator and denominator of a fraction in simplest form that is equivalent to

Mathematics
2 answers:
Olenka [21]3 years ago
7 0
\frac{941}{999}
Romashka [77]3 years ago
4 0
94/100= 47/50, I'm not sure which number your teacher wanted you to round to so I just did the normal first two after dot
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Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teache
Tju [1.3M]

Answer:

The 95% confidence interval would be given (0.0109;0.1651).  

We are confident at 95% that the difference between the two proportions is between 0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651

Step-by-step explanation:

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_A represent the real population proportion for elementary school  

\hat p_A =\frac{226}{397}=0.569 represent the estimated proportion for elementary school

n_A=397 is the sample size required for Brand A

p_B represent the real population proportion for high school teachers  

\hat p_B =\frac{129}{268}=0.481 represent the estimated proportion for high school teachers

n_B=268 is the sample size required for Brand B

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_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

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:  

(0.569-0.481) - 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.0109  

(0.569-0.481) + 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.1651  

And the 95% confidence interval would be given (0.0109;0.1651).  

We are confident at 95% that the difference between the two proportions is between 0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651

5 0
2 years ago
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