Respuesta :
Answer:
99%, confidence interval for the true mean lifespan of this product is [12.08 years , 13.92 years].
Step-by-step explanation:
We are given that a toy maker claims his best product has an average lifespan of exactly 18 years.
The product evaluator was provided data collected from a random sample of 35 people who used the product. Using the data, an average product lifespan of 13 years and a standard deviation of 2 years was calculated.
Firstly, the pivotal quantity for 99% 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 product lifespan = 13 years
n = sample of people = 35
s = sample standard deviation = 2 years
[tex]\mu[/tex] = true mean lifespan
Here for constructing 99% confidence interval we have used One-sample t statistics because we don't know about the population standard deviation.
So, 99% confidence interval for the true mean, [tex]\mu[/tex] is ;
P(-2.728 < [tex]t_3_4[/tex] < 2.728) = 0.99 {As the critical value of t at 34 degree
of freedom are -2.728 & 2.728 with P = 0.5%}
P(-2.728 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.728) = 0.99
P( [tex]-2.728 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.728 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-2.728 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.728 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.728 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.728 \times {\frac{s}{\sqrt{n} } }[/tex]]
= [ [tex]13-2.728 \times {\frac{2}{\sqrt{35} } }[/tex] , [tex]13+2.728 \times {\frac{2}{\sqrt{35} } }[/tex] ]
= [12.08 , 13.92]
Therefore, 99% confidence interval for the true mean lifespan of this product is [12.08 years , 13.92 years].
