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In 2006, 95% of new cars in the US came with a spare tire. In 2017, 72% came with a spare tire.
Suppose a researcher takes separate random samples of 250 cars made in each year, then looks at the
difference between the sample proportions P2006 - P2017 of cars with spare tires in each sample.
What are the mean and standard deviation of the sampling distribution of P2006 - P2017?

Choose 1 answer:

A. mean of P2006-P2017: 57.5

Standard Deviation of P2006-2017: Square root of .95(.05)/250 + .72(.28)/250


B C and D in picture

In 2006 95 of new cars in the US came with a spare tire In 2017 72 came with a spare tire Suppose a researcher takes separate random samples of 250 cars made in class=

Respuesta :

Using the Central Limit Theorem, it is found that:

  • The mean is 0.23.
  • The standard deviation is [tex]s = \sqrt{\frac{0.95(0.05)}{250} + \frac{0.72(0.28)}{250}}[/tex].

Central Limit Theorem

  • It states that for a proportion p in a sample of size n, the sampling distribution of sample proportions has mean [tex]p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]
  • When two variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In 2006, 95% of new cars in the US came with a spare tire, with a sample of 250, hence:

[tex]p_1 = 0.95, s_1 = \sqrt{\frac{0.95(0.05)}{250}}[/tex]

In 2017, 72% of new cars in the US came with a spare tire, with a sample of 250, hence:

[tex]p_2 = 0.72, s_2 = \sqrt{\frac{0.72(0.28)}{250}}[/tex]

Hence, for the distribution of differences:

[tex]p = p_1 - p_2 = 0.95 - 0.72 = 0.23[/tex]

[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{\frac{0.95(0.05)}{250} + \frac{0.72(0.28)}{250}}[/tex]

To learn more about the Central Limit Theorem, you can take a look at https://brainly.com/question/16695444

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