Question

A sample of students that had stuffy or runny noses were blindly given a single Skittles candy to put in their mouth. They were told the five possible flavors and then were asked which flavor they had. Of the 118 students tested, 44 gave the correct answer. Find a 95% confidence interval for the proportion of all students with stuffy or runny noses that could correctly identify a Skittles flavor blindly.

Answer 1

The confidence interval is
`0.2857<  p <  0.4601`

Explanation:

From the question we are told that

The sample size is n = 118

The number that gave the correct answer is k = 44

Generally the sample proportion is mathematically represented as

`\^ p = (44)/(118)`

=>
`\^ p =0.3729`

Generally given that the confidence level is 95% the level of significance is mathematically represented as

`\alpha = (100 -95) \%`

=>
`\alpha = 0.05`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E =  Z_{(\alpha )/(2) } * \sqrt{(p(1 - p))/(n) }`

=>
`E =  1.96 * \sqrt{(0.3729(1 - 0.3729))/(118) }`

=>
`E = 0.0872`

Generally 95% confidence interval is mathematically represented as

`\r p -E <  p <  \r p +E`

=>
`0.3729 -0.0872<  p <  0.3729 +0.0872`

=>
`0.2857<  p <  0.4601`