Question

Obesity is seen as a growing health problem in many countries, including the United States. To determine whether or not there has been an increase in obesity in men over the past ten years, medical records for 50 randomly sampled men from the year 2000 and for 75 randomly sampled men from the year 2010 were analyzed. Out of the 50 men from 2000, 10 were assigned as obese according to their height and weight while 30 out of the 75 men from 2010 were assigned as obese.

The 90% confidence interval for the difference in the proportion of obese men between the years 2000 and 2010 is (-0.3316, -0.0684). Select the correct interpretation of this confidence interval.
a. The proportion of men who were obese in the year 2000 is between 0.0684 and 0.3316 less than the proportion of men who were obese in the year 2010.b. We are 90% confident that the proportion of men who were obese in the year 2000 is between 0.0684 and 0.3316 less than the proportion of men who were obese in the year 2010.c. We are 90% confident that the proportion of men who were obese in the year 2000 is between 0.0684 and 0.3316 greater than the proportion of men who were obese in the year 2010.d. In 90% of all samples, the proportion of men who were obese in the year 2000 is between 0.0684 and 0.3316 less than the proportion of men who were obese in the year 2010.

Answer 1

`(0.2-0.4) - 1.64 \sqrt{(0.2(1-0.2))/(50) +(0.4(1-0.4))/(75)}=-0.3316`

`(0.2-0.4) + 1.64 \sqrt{(0.2(1-0.2))/(50) +(0.4(1-0.4))/(75)}=-0.0684`

And the 90% confidence interval would be given (-0.3316;-0.0684).

b. We are 90% confident that the proportion of men who were obese in the year 2000 is between 0.0684 and 0.3316 less than the proportion of men who were obese in the year 2010.

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 2000

`\hat p_A =(10)/(50)=0.2` represent the estimated proportion for 2000

`n_A=50` is the sample size required for 2000

`p_B` represent the real population proportion for 2010

`\hat p_B =(30)/(75)=0.4` represent the estimated proportion for 2010

`n_B=75` is the sample size required for 2010

`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=1-0.9=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:

`(0.2-0.4) - 1.64 \sqrt{(0.2(1-0.2))/(50) +(0.4(1-0.4))/(75)}=-0.3316`

`(0.2-0.4) + 1.64 \sqrt{(0.2(1-0.2))/(50) +(0.4(1-0.4))/(75)}=-0.0684`

And the 90% confidence interval would be given (-0.3316;-0.0684).

We are confident at 90% that the difference between the two proportions is between
`-0.3316 \leq p_A -p_B \leq -0.0684`

The correct interpretation would be:

b. We are 90% confident that the proportion of men who were obese in the year 2000 is between 0.0684 and 0.3316 less than the proportion of men who were obese in the year 2010.