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In 2020 researchers took random samples to compare the proportion of adult females to the proportion of adult males who use 2 or more credit/debit cards. The adults were aged 21 and older and they lived in all areas of the U.S. Based on the data collected, the researchers computed a 96 percent confidence interval for the true difference between the proportions of adult females minus the proportion of adult males who use 2 or more credit/debit cards. The confidence interval is: (0.013, 0.049). Assume all checks have been verified.
Question:
Choose the best answer below concerning the confidence interval for the difference between the true proportions of adult females minus the true proportion of adult males who use 2 or more credit/debit cards.

In 2020 researchers took random samples to compare the proportion of adult females to the proportion of adult males who use 2 or more creditdebit cards The adul class=

Respuesta :

Using the z-distribution, it is found that the correct option is:

[tex]p_1 - p_2 \pm z\sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}[/tex]

A confidence interval is given by:

[tex]p \pm zs[/tex]

In which:

  • p is the estimate of the proportion.
  • z is the critical value.
  • s is the standard error.

In this problem, the proportions are:

  • A proportion [tex]p_1[/tex] in a sample of size [tex]n_1[/tex].
  • [tex]p_2[/tex] in a sample of size [tex]n_2[/tex].

Hence, the standard errors are:

[tex]s_1 = \sqrt{\frac{p_1(1 - p_1)}{n_1}}, s_2 = \sqrt{\frac{p_2(1 - p_2)}{n_2}}[/tex]

When two proportions are subtracted, the estimate is the subtraction of proportions, while the standard error of the subtraction is the square root of the sum of each standard error squared, hence:

[tex]p = p_1 - p_2[/tex]

[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}[/tex]

Thus, the interval is:

[tex]p_1 - p_2 \pm z\sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}[/tex]

A similar problem is given at https://brainly.com/question/12517818

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