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Suppose you take an SRS of size 1000 of Connecticut students eligible to take the SAT and find that 85% plan to take the SAT during the 2010-2011 school year. Under the null hypothesis H0: p = 0.81, the sampling distribution has mean = _____ and standard deviation = 0.0124. Answer to two decimal places.

Respuesta :

Answer:

[tex]\mu_{p}=p=0.81[/tex]

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

So under the null hypothesis the mean for the population proportion is p

[tex]\mu_{p}=p=0.81[/tex]

And the standard deviationis given by:

[tex]\sigma_{p}=\sqrt{\frac{p_0(1-p_o)}{n}}=\sqrt{\frac{0.81(1-0.81)}{1000}}=0.0124[/tex]

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