Mathematics: 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?

In a recent survey of county high school students,

Ответ:

unkn03n78

13.01.2022

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.

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".

p_A represent the real population proportion for males

$\hat p_A =\frac{66}{100}=0.66$ 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

$\hat p_B =\frac{47}{100}=0.47$ represent the estimated proportion for females

n_B=100 is the sample size required for females

represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

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

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

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:

$(0.66-0.47) - 1.64 \sqrt{\frac{0.66(1-0.66)}{100} +\frac{0.47(1-0.47)}{100}}=0.0771$

$(0.66-0.47) + 1.64 \sqrt{\frac{0.66(1-0.66)}{100} +\frac{0.47(1-0.47)}{100}}=0.30329$

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.

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