Respuesta :
Answer:
a) [tex](0.67-0.56) - 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.054[/tex]
[tex](0.67-0.56) + 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.166[/tex]
And the 90% confidence interval would be given (0.054;0.166).
b) [tex](0.31-0.25) - 1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=-0.0122[/tex]
[tex](0.31-0.25) +1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=0.132[/tex]
And the 90% confidence interval would be given (-0.0122;0.132).
c) [tex](0.46-0.61) - 1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.260[/tex]
[tex](0.46-0.61) +1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.0404[/tex]
And the 90% confidence interval would be given (-0.260;-0.0404).
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".
Part a
[tex]p_1[/tex] represent the real population proportion for sample 1
[tex]\hat p_1 =0.67[/tex] represent the estimated proportion for sample 1
[tex]n_A=400[/tex] is the sample size required for sample 1
[tex]p_2[/tex] represent the real population proportion for sample 2
[tex]\hat p_2 =0.56[/tex] represent the estimated proportion for sample 2
[tex]n_2=400[/tex] is the sample size required for sample 2
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]
For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex](0.67-0.56) - 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.054[/tex]
[tex](0.67-0.56) + 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.166[/tex]
And the 90% confidence interval would be given (0.054;0.166).
Part b
[tex](0.31-0.25) - 1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=-0.0122[/tex]
[tex](0.31-0.25) +1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=0.132[/tex]
And the 90% confidence interval would be given (-0.0122;0.132).
Part c
[tex](0.46-0.61) - 1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.260[/tex]
[tex](0.46-0.61) +1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.0404[/tex]
And the 90% confidence interval would be given (-0.260;-0.0404).
