A point estimate for the population proportion will be found using the sample data. A survey of 358 working parents was conducted and it was found that 250 said they spend too little time with their children because of work commitments.

(b) At 95% confidence, what is the margin of error? Recall the formula for the margin of error, E, given below where z , is the z value corresponding to an upper tail area of _ and a confidence level of 1 - a, p is the point estimate for the population proportion, and n is the sample size. E = Z Q/2\\ p(1 - p) There were 358 adults in the survey, so n = , and the sample proportion was found to be p = 0.6983. Now the value of z , is needed. A margin of error for 95% confidence is needed. Common values for z , for various confidence levels are given below. Confidence Level N R Z a / 2 90% 0.10 0.05 1.645 95% 0.05 0.025 1.960 99% 0.01 0.005 2.576 According to the table, for a 95% confidence level, the necessary value for za/2 is(3) Using the sample data, what is the margin of error in dollars assodated with a 95% condence interval? Since the population standard deviation is known, the normal distribution will be used in the calculation of the margin of error, E. The value 0' According to the table, for a 95% condence level, the necessary value for zm,2 is E=z tr 41.32 H 2m,2 is the 2 value corresponding to an upper tail area of Condence Level (I ; z": 90% 0.10 0.05 1.645 95% 0.05 0.025 1.960 99% 0.01 0.005 2.576 E and a condence level of 1 a. Common values for z 2 \"[2 for various condence levels are given below. The survey was conducted among a sample of 61 discount brokers, and we are to assume a known population standard deviation of $14. Therefore, (I = and :r =