Respuesta :
Answer:
a) The 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day is (1.49, 1.57).
d) The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for any individual is between 1.49 and 1.57 hours.
(c and b can not be concluded from the confidence interval)
Step-by-step explanation:
We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=1.53.
The sample size is N=1082.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.71}{\sqrt{1082}}=\dfrac{0.71}{32.89}=0.022[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=1082-1=1081[/tex]
The t-value for a 95% confidence interval and 1081 degrees of freedom is t=1.962.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=1.962 \cdot 0.022=0.042[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M-t \cdot s_M = 1.53-0.042=1.49\\\\UL=M+t \cdot s_M = 1.53+0.042=1.57[/tex][tex]LL=M-t \cdot s_M = 1.53-0.042=1.49\\\\UL=M+t \cdot s_M = 1.53+0.042=1.57[/tex]
The 95% confidence interval for the mean is (1.49, 1.57).
