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

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

Step-by-step 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. =    ~

where, = sample average rating = 5.6

            s = sample standard deviation = 4.6

            n = sample of tobacco smokers = 81

            = 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, is ;

P(-1.993 < < 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 < < 1.993) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                                            = [ , ]

                                            = [4.6 , 6.6]

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

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