Question

In a poll of 555 randomly selected students, 40% stated that they enjoyed statisticsa. Identify the number of students who say that they enjoy statistics? Round to the nearest whole student if necessary.b. Construct a 95% confidence interval estimate of the percentage of all students who say that they enjoy statistics.c. Can we safely conclude that majority of students enjoy statistics? Explain.

Answer 1

a

`N = 222`

b

`0.36< p <0.44`

c

No we can not safely conclude that majority of students enjoy statistics because the upper limit of the confidence level is less than 50%

Explanation:

From the question we are told that

The sample size is
`n = 555`

The percentage that enjoyed statistics is
`\r p = 40% = 0.40`

Generally the number of student who say they enjoyed statistics is mathematically represented as

`N = 0.40 * 55`

=>
`N = 222`

Given that the confidence level is 95% then the level of significance is mathematically represented as

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

=>
`\alpha = 0.05`

The critical value of
`(\alpha )/(2)` obtained from the normal distribution table is
`Z_{(\alpha )/(2) } = 1.96`

The margin of error is mathematically represented as

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

=>
`E = 1.96 * \sqrt{ (0.40 (1- 0.40 ))/( 555) }`

=>.
`E = 0.04`

The 95% confidence interval is mathematically represented as

`\r p - E < p < \r p + E`

=>
`0.40 - 0.04 < p <0.40 + 0.04`

=>
`0.36< p <0.44`

No we can not safely conclude that majority of students enjoy statistics because the upper limit of the confidence level is less than 50%