a survey of 504 citizens found that 378 of them favor a new bill introduced by the city. We want to find a 95% confidence interv
1 answer:
The lower limit of the interval is 0.712 if the survey of 504 citizens found that 378 of them favor a new bill introduced by the city.
<h3>What is a confidence interval?</h3>It is defined as the sampling distribution following an approximately normal distribution for known standard deviation.
We have:
A survey of 504 citizens found that 378 of them favor a new bill introduced by the city.
Sample proportion = p = 378/504 = 0.75
q = 1 - p = 1 - 0.75 = 0.25
SD = 0.01928
For 95% confidence interval Z value = 1.96
Lower limit = 0.75 - 1.96(0.01928)
= 0.712
Thus, the lower limit of the interval is 0.712 if the survey of 504 citizens found that 378 of them favor a new bill introduced by the city.
Learn more about the confidence interval here:
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Answer:
The total minutes for last month is 300 + 800 = 1100 minutes.
Step-by-step explanation:
Last month a phone company charged $ 0.07 per minute for the first 300 minutes of calling time and $0.05 per minute for each minute of calling time in excess of the first 300 minutes . For how many minutes of calling time was Bob charged by the phone company last month?1) Last month the phone company charged Bob $21.00 for his calling time that was charged at $0.07 per minute.2) Last month the phone company charged Bob $40.00 for his calling time that was charged at $0.05 per minute.
1) If Bob was charged $0.07 per minute, it means the call was within the first 300 minutes. He was charged $21.00 for last month at $0.07per minute. Therefore, he called for:
He called for 300 minutes with $21.00
2) If Bob was charged $0.05 per minute, it means the call was outside the first 300 minutes. He was charged $40.00 for last month at $0.05 per minute. Therefore, he called for:
He called for additional 800 minutes with $40
Therefore, the total minutes for last month is 300 + 800 = 1100 minutes.