Question

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Houston. The study was performed at the end of the school year and asked each if they had used steroids at any point in the last school year. Of the upperclassmen, 25 claimed to have used steroids in the last school year, and of the underclassmen, 19 claimed to have used steroids. Run a 95% confidence interval to test for a significant difference in the proportions of students who used steroids.

Answer 1

`-0.048<P_1-P_2<0.036`

Do not reject
`H_0:P_1-P_2=0`

Explanation:

From the question we are told that

Sample size
`n_1=350`

Sample size
`n_2=250`

Sample proportion 1
`\hat P= (25)/(350) =>0.07`

Sample proportion 2
`\hat P= (19)/(250) =>0.076`

95% confidence interval

Generally for 95% confidence level

Level of significance

`\alpha = 1-0.95=>0.05`

`\alpha /2=(0.05)/(2) =>0.025`

Therefore

`Z_a_/_2=1.96`

Generally the equation for confidence interval between
`P_1 - P_2` is mathematically given as

`(\hat P_1-\hat P_2)\pm Z_a_/_2\sqrt{(\hat P_1(1-\hat P_1))/(n_1)+(\hat P_2(1-\hat P_2))/(n_2) }`

`(0.07-0.076)\pm 1.96\sqrt{(0.07(1-0.07))/(350)+(0.076(1-0.076))/(250) }`

`(0.07-0.076)\pm 1.96√(4.66896*10^-^4 )`

`(-0.006)\pm 0.042`

`(-0.006)- 0.042=>-0.048`

`(-0.006)+ 0.042=>0.036`

Therefore

Confidence interval is

`-0.048<P_1-P_2<0.036`

Conclusion

Given the confidence interval has zero

Therefore do not reject
`H_0:P_1-P_2=0`