Question

A sample of 81 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10, with 10 corresponding to "completely effective" and 1 corresponding to "completely ineffective". The average rating was 5.6 and the standard deviation was 4.6. Construct a 95% confidence interval for the mean score. 5.2 < μ < 6.0 0 < μ < 5.6 4.6 < μ < 6.6 5.1 < μ < 6.1

Answer 1

95% confidence interval for the true mean score is [4.6 , 6.6].

Explanation:

We are given that a sample of 81 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10.

The average rating was 5.6 and the standard deviation was 4.6.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample average rating = 5.6

s = sample standard deviation = 4.6

n = sample of tobacco smokers = 81

`\mu` = population mean score

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean score,
`\mu` is ;

P(-1.993 <
`t_8_0` < 1.993) = 0.95 {As the critical value of t at 80 degree of

freedom are -1.993 & 1.993 with P = 2.5%}

P(-1.993 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 1.993) = 0.95

P(
`-1.993 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.993 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-1.993 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+1.993 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-1.993 * {(s)/(√(n) ) }` ,
`\bar X+1.993 * {(s)/(√(n) ) }` ]

= [
`5.6-1.993 * {(4.6)/(√(81) ) }` ,
`5.6+1.993 * {(4.6)/(√(81) ) }` ]

= [4.6 , 6.6]

Therefore, 95% confidence interval for the true mean score is [4.6 , 6.6].