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A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let ^ p denote the proportion in the sample who say they support the increase. Suppose that 13% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is

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Answer:

The standard deviation of the sampling distribution is [tex]\sigma =0.0087[/tex]

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  1500

     The population proportion is  [tex]p = 0.013[/tex]

Generally the standard deviation of this sampling distribution is mathematically represented as

      [tex]\sigma = \sqrt{ \frac{p (1- p)}{ n} }[/tex]

=>     [tex]\sigma = \sqrt{\frac{0.13(1-0.13)}{1500} }[/tex]

=>   [tex]\sigma =0.0087[/tex]

     

The standard deviation of the considered sampling distribution of the proportion supporting the increase is 0.0027 approx.

How to find the sample standard deviation for distribution of a sample proportion?

Suppose the sample proportion be denoted by [tex]\hat{p}[/tex], then, its distribution is normally distributed with mean [tex]p[/tex] and the standard deviation of distribution of [tex]\hat{p}[/tex] is given by

 [tex]\sigma = \sqrt{\dfrac{p(1-p)}{n}}[/tex]

where n is the sample size. It is true until [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex]

For the given case, we're given that:

  • Sample size=  n = 10
  • Proportion average value = p = 13% = 0.13
  • Standard deviation = [tex]\sigma = \sqrt{\dfrac{0.13(1-0.13)}{1500}} = \sqrt{\dfrac{0.13 \times 0.87}{1500}} \approx 0.0027[/tex]

Thus, standard deviation of the sampling distribution is approximately 0.0027

Learn more about standard deviation of proportion distribution here:

https://brainly.com/question/16548769

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