Answer: The sample should contain 4870 cans of paint.
Step-by-step explanation:
According to the given information , we have
The prior estimate of proportion= [tex]p=0.10[/tex]
Margin of error : E=0.01
Critical value for 98% confidence interval =[tex]z_{\alpha/2}=2.326[/tex]
Formula to find the sample size :
[tex]n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2[/tex]
i.e. [tex]n=0.10(1-0.10)(\dfrac{2.326}{0.01})^2[/tex]
[tex]=4869.2484\approx4870[/tex]
Hence, the sample should contain 4870 cans of paint.