Question

In a recent survey of county high school students, 100 males and 100 females, 66 of the male students and 47 of the female students sampled admitted that they consumed alcohol on a regular basis. Find a 90% confidence interval for the difference between the proportion of male and female students that consume alcohol on a regular basis. Can you draw any conclusions from the confidence interval?

Answer 1

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
`0.0771 \leq p_(Males) -p_(Females) \leq 0.30329`

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.

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".

`p_A` represent the real population proportion for males

`\hat p_A =(66)/(100)=0.66` represent the estimated proportion for males

`n_A=100` is the sample size required for males

`p_B` represent the real population proportion for females

`\hat p_B =(47)/(100)=0.47` represent the estimated proportion for females

`n_B=100` is the sample size required for females

`z` represent the critical value for the margin of error

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

For the 90% confidence interval the value of
`\alpha=1-0.90=0.1` and
`\alpha/2=0.05`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.64`

And replacing into the confidence interval formula we got:

`(0.66-0.47) - 1.64 \sqrt{(0.66(1-0.66))/(100) +(0.47(1-0.47))/(100)}=0.0771`

`(0.66-0.47) + 1.64 \sqrt{(0.66(1-0.66))/(100) +(0.47(1-0.47))/(100)}=0.30329`

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
`0.0771 \leq p_(Males) -p_(Females) \leq 0.30329`

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.