Respuesta :
Answer:
The 90% confidence interval would be given (0.0771;0.30329).
We are confident at 90% that the difference between the two proportions is between [tex]0.0771 \leq p_{Males} -p_{Females} \leq 0.30329[/tex]
Can you draw any conclusions from the confidence interval?
Yes since the confidence interval not contains the 0 we can conclude that the proportion of males is significantly higher than the proportion of females at 10 % of significance.
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".
[tex]p_A[/tex] represent the real population proportion for males
[tex]\hat p_A =\frac{66}{100}=0.66[/tex] represent the estimated proportion for males
[tex]n_A=100[/tex] is the sample size required for males
[tex]p_B[/tex] represent the real population proportion for females
[tex]\hat p_B =\frac{47}{100}=0.47[/tex] represent the estimated proportion for females
[tex]n_B=100[/tex] is the sample size required for females
[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_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
For the 90% confidence interval the value of [tex]\alpha=1-0.90=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](0.66-0.47) - 1.64 \sqrt{\frac{0.66(1-0.66)}{100} +\frac{0.47(1-0.47)}{100}}=0.0771[/tex]
[tex](0.66-0.47) + 1.64 \sqrt{\frac{0.66(1-0.66)}{100} +\frac{0.47(1-0.47)}{100}}=0.30329[/tex]
And the 90% confidence interval would be given (0.0771;0.30329).
We are confident at 90% that the difference between the two proportions is between [tex]0.0771 \leq p_{Males} -p_{Females} \leq 0.30329[/tex]
Can you draw any conclusions from the confidence interval?
Yes since the confidence interval not contains the 0 we can conclude that the proportion of males is significantly higher than the proportion of females at 10 % of significance.
