Asurvey of 85 families showed that 36 owned at least one dvd player. find the 99% confidence interval estimate of the true proportion of families who own at least one dvd player. place your limits, rounded to 3 decimal places, in the blanks. place the lower limit in the first blank

Step-by-step explanation:

The confidence interval for population mean is given by :-

, where is the sample proportion, n is the sample size , is the critical z-value.

Given : Significance level :

Sample size : n= 85

Critical value :

Sample proportion:

Now, the  99% confidence level will be :

Hence, the  99% confidence interval estimate of the true proportion of families who own at least one DVD player is  

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation:

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step-by-step explanation: