Answer: (0.665 , 0.777)
Step-by-step explanation:
Given : Sample size of consumers: n=420 ;
Number of consumers say their ColorSmart-5000 television sets did not need a repair=303
The the sample proportion for consumers say their ColorSmart-5000 television sets did not need a repair : [tex]\hat{p}=\dfrac{303}{420}\approx0.721[/tex]
Significance level : [tex]\alpha:1-0.99=0.01[/tex]
Critical value = [tex]z_{\alpha/2}=\pm2.576[/tex]
The confidence interval for population proportion is given by :_
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]
i.e. [tex]0.721\pm (2.576)\sqrt{\dfrac{0.721(1-0.721)}{420}}[/tex]
[tex]\approx0.721\pm 0.056=(0.721-0.056,0.721+0.056)\\\\=(0.665\ ,\ 0.777)[/tex]
Hence, a 99 percent confidence interval for the proportion of all ColorSmart-5000 television sets that have lasted at least five years without needing a single repair = (0.665 , 0.777)