How Many Times Do You Go Pee In One Day?
By: Olivia M & Makenna M
For our chapter seven project, Olivia and Makenna collected data regarding the number of times someone urinates in one day. We randomly selected individuals in each class, and asked them in person for a simple random sample. Then we collected all of our data and plugged it into a spreadsheet. After collecting all of our data we calculated one variable statistics and came up with a mean of 5.5. Which meant that most of the people we surveyed went pee approximately five to six times in one day. From the one variable stats we also found out that the sample standard deviation is 2.43, which meant that our data is spread out by 2.43. Our five number summary is minimum of one, Q1 is four, median is five, Q3 is six, and our maximum is fourteen.
To determine if our data is normal we looked at our google spreadsheet. Our histogram looked generally bell shaped, although there were much taller bars as well as a lot of short bars. Since our histogram looked generally bell shaped, we then checked for outliers. By multiplying the interquartile range by 1.5 we got three, then you either subtract or add three to the Q1 and Q3, anything above or below that is an outlier. So anything below one is an outlier, and anything above nine is also an outlier. So in anyone peed less than one or more than nine times in one day, they were considered an outlier. We technically had three outliers: ten, twelve, and fourteen, so our data wouldn’t be considered normal, but for this project we assumed it was normal. Also to check for skewness we used Pearson's Index. We took 3*(5.5-5)/ 2.43 and got 0.617, which is in between -1 and 1. So to calculate if our data is normal, it meets the criteria of bell shaped histogram and skewness, but not outliers because we had three when you are only supposed to have one.
The empirical rule is calculated using the mean and normal distribution curve. Our mean was 5.5 and the standard deviation was 2.43. Calculating standard deviations of a normal curve, you add or subtract the standard deviation to the mean. For example, two standard deviations away which is 95% would be between the values 0.64 and 10.36 on a normal curve. This means that 95% of people that we surveyed pee between one to ten times in one day. One standard deviation away is 68% which equaled 3.07 to 7.93, meaning that 68% of the people we surveyed peed three to eight times in one day. Three standard deviations is 99.7% equaled -1.79 to 12.79, this means once again that 99.7% of the people we surveyed would generally pee zero to thirteen times. Since someone cannot pee a negative number of times, we rounded up to zero. Almost all of our data is covered in the 99.7% because only 3% of our data pee more than thirteen times in one day.
For the number of times we urinate in a day, Makenna answered 4 and Olivia answered 7. The probability that someone would answer less than or equal to four would be 0.7324. This would mean that someone peed either four times or less. This can be calculated by taking our x of four and subtracting it by 5.5 then dividing it by the standard deviation of 2.43. From there we got a z score of -0.6172, then we used our normal distribution z table to find the probability of randomly selecting a value less than or equal to four or four times or less that someone pees. The table gave us 0.2676, which is a right tail probability so we subtracted it from one and ended up with 0.7324. We followed the same steps to find the probability greater than seven or seven or more times that someone pees, except we didn’t have to subtract the probability by one because it was already a left tail. The probabilities that someone peed less than four times or greater than seven times in one day were exactly the same value because they were the same distance away from our mean. The two probabilities were the same so finding a value between four and seven would be zero when the two probabilities 0.7324 were subtracted.
We had to use the z-scores to find the probability that 30 people would average greater than our personal response of four or seven times we pee in one day. To find this, we used the central limit theorem. We took x minus the mean divided by standard deviation divided by square root of the sample size. In numbers, 7-5.5/2.43/30 which equals 3.3810. The probability of picking 30 people peeing greater than seven times in one day would be 0.0004. So less than one person would pee more than seven times when randomly selecting 30 data values. Then we did it again to find the probability less than 4 times someone pees, once again was the same except value but negative. That means that about 0.0004 people would pee less than four times in a random sample of thirty people. To find the probability between responses of less than four and greater than seven, you have to subtract the two 0.0004-0.0004 which equals 0. Which means that almost all of our data would fall between four and seven times peeing in one day.
We had to find the raw data value that was below 3% of where our data fell. To find this we did x equals z times standard deviation plus mean, which actually is -1.88(2.43)+5.5=0.9316. Which means that only 3% of people pee less than one time a day. Then we did it again to find the raw data which only 3% of data would fall and that ended up being 10.0684. Which means that 3% of people pee more than ten times in one day.
To determine if our data is normal we looked at our google spreadsheet. Our histogram looked generally bell shaped, although there were much taller bars as well as a lot of short bars. Since our histogram looked generally bell shaped, we then checked for outliers. By multiplying the interquartile range by 1.5 we got three, then you either subtract or add three to the Q1 and Q3, anything above or below that is an outlier. So anything below one is an outlier, and anything above nine is also an outlier. So in anyone peed less than one or more than nine times in one day, they were considered an outlier. We technically had three outliers: ten, twelve, and fourteen, so our data wouldn’t be considered normal, but for this project we assumed it was normal. Also to check for skewness we used Pearson's Index. We took 3*(5.5-5)/ 2.43 and got 0.617, which is in between -1 and 1. So to calculate if our data is normal, it meets the criteria of bell shaped histogram and skewness, but not outliers because we had three when you are only supposed to have one.
The empirical rule is calculated using the mean and normal distribution curve. Our mean was 5.5 and the standard deviation was 2.43. Calculating standard deviations of a normal curve, you add or subtract the standard deviation to the mean. For example, two standard deviations away which is 95% would be between the values 0.64 and 10.36 on a normal curve. This means that 95% of people that we surveyed pee between one to ten times in one day. One standard deviation away is 68% which equaled 3.07 to 7.93, meaning that 68% of the people we surveyed peed three to eight times in one day. Three standard deviations is 99.7% equaled -1.79 to 12.79, this means once again that 99.7% of the people we surveyed would generally pee zero to thirteen times. Since someone cannot pee a negative number of times, we rounded up to zero. Almost all of our data is covered in the 99.7% because only 3% of our data pee more than thirteen times in one day.
For the number of times we urinate in a day, Makenna answered 4 and Olivia answered 7. The probability that someone would answer less than or equal to four would be 0.7324. This would mean that someone peed either four times or less. This can be calculated by taking our x of four and subtracting it by 5.5 then dividing it by the standard deviation of 2.43. From there we got a z score of -0.6172, then we used our normal distribution z table to find the probability of randomly selecting a value less than or equal to four or four times or less that someone pees. The table gave us 0.2676, which is a right tail probability so we subtracted it from one and ended up with 0.7324. We followed the same steps to find the probability greater than seven or seven or more times that someone pees, except we didn’t have to subtract the probability by one because it was already a left tail. The probabilities that someone peed less than four times or greater than seven times in one day were exactly the same value because they were the same distance away from our mean. The two probabilities were the same so finding a value between four and seven would be zero when the two probabilities 0.7324 were subtracted.
We had to use the z-scores to find the probability that 30 people would average greater than our personal response of four or seven times we pee in one day. To find this, we used the central limit theorem. We took x minus the mean divided by standard deviation divided by square root of the sample size. In numbers, 7-5.5/2.43/30 which equals 3.3810. The probability of picking 30 people peeing greater than seven times in one day would be 0.0004. So less than one person would pee more than seven times when randomly selecting 30 data values. Then we did it again to find the probability less than 4 times someone pees, once again was the same except value but negative. That means that about 0.0004 people would pee less than four times in a random sample of thirty people. To find the probability between responses of less than four and greater than seven, you have to subtract the two 0.0004-0.0004 which equals 0. Which means that almost all of our data would fall between four and seven times peeing in one day.
We had to find the raw data value that was below 3% of where our data fell. To find this we did x equals z times standard deviation plus mean, which actually is -1.88(2.43)+5.5=0.9316. Which means that only 3% of people pee less than one time a day. Then we did it again to find the raw data which only 3% of data would fall and that ended up being 10.0684. Which means that 3% of people pee more than ten times in one day.