Answer: (0.79, 3.05 )
Step-by-step explanation:
Confidence interval for the population mean:
[tex]\overline{x}\pm z^c\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\sigma[/tex] = population standard deviation
n= sample size.
[tex]z^c[/tex] = Critical z value for confidence interval c.
As per given:
n= 10
[tex]\overline{x}=1.92[/tex]
[tex]\sigma=1.83[/tex]
Critical z-value for 95% confidence = 1.96
A 95% confidence interval for the mean opinion in the population of all licensed drivers:-
[tex]1.92\pm (1.96)\dfrac{1.83}{\sqrt{10}}\\\\=1.92\pm1.134\\\\= (1.92-1.134,\ 1.92+1.134)\\\\\approx(0.79,\ 3.05 )[/tex]
Hence, a 95% confidence interval for the mean opinion in the population of all licensed drivers = (0.79, 3.05)
