A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x= 14.8 Books & S= 16.6 books. construct a 95% confidence interval for the mean number of books read. Interpret the interval.
construct a 95% confidence interval for the mean number of books people read and interpret the results. Select the correct choice below and fill in the answer boxes to complete your choice.
a) if repeater samples are taken, 95% of them will have a sample mean between _and .
b) there is a 95% chance that the true me number of books read is between and .c) there is 95% confidence that the population mean number of books read is between __ and _.

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=14.8.

The sample size is N=1003.

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{16.6}{\sqrt{1003}}=\dfrac{16.6}{31.67}=0.524[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=1003-1=1002[/tex]

The t-value for a 95% confidence interval and 1002 degrees of freedom is t=1.96.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_M=1.96 \cdot 0.524=1.03[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M-t \cdot s_M = 14.8-1.03=13.77\\\\UL=M+t \cdot s_M = 14.8+1.03=15.83[/tex]

The 95% confidence interval for the mean number of books read is (13.77, 15.83).

This indicates that there is 95% confidence that the true mean is within 13.77 and 15.83. Also, that if we take multiples samples, it is expected that 95% of the sample means will fall within this interval.