Answer :
Answer:
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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average rating = 5.6
s = sample standard deviation = 4.6
n = sample of tobacco smokers = 81
[tex]\mu[/tex] = 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, [tex]\mu[/tex] is ;
P(-1.993 < [tex]t_8_0[/tex] < 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 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.993) = 0.95
P( [tex]-1.993 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.993 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.993 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.993 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.993 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.993 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]5.6-1.993 \times {\frac{4.6}{\sqrt{81} } }[/tex] , [tex]5.6+1.993 \times {\frac{4.6}{\sqrt{81} } }[/tex] ]
= [4.6 , 6.6]
Therefore, 95% confidence interval for the true mean score is [4.6 , 6.6].