I have given you the information from 2-4 in hopes to help with the answer for #6.. Thank you!
2. Calculate the mean and standard deviation of the probability distribution created by rolling a die. Either show work or explain how your answer was calculated.
x 1 2 3 4 5 6 sum
p(x) 1/6 1/6 1/6 1/6 1/6 1/6 6/6 = 1
x p(x) 1/6 2/6 3/6 4/6 5/6 6/6 21/6 = 3.5
x^2 p(x) 1/6 4/6 9/6 16/6 25/6 36/6 91/6 = 15.1667
Mean: ___3.5_________ Standard deviation: is the sq root of the variance_= The variance is 91/6 - (7/2)^2 = 35/12 = 2.916666... = 1.7078________
3. Give the mean for the mean column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
The Mean for Column “mean” is 3.56 It is very close to the parameter of interest but is not equal to it. You could calculate a confidence interval for the mean of the mean column, but a specific confidence interval would need to be provided. In that case, the confidence interval would be centered on 3.56, not 3.5.
4. Give the mean for the median column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
The Mean for the Median column is 3.6 The mean here is close to the mean in question 2, but is not as close as the answer from question 3.
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5. Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
Standard deviation of Mean = 0.4762
Standard deviation of Median = 0.7539
The Standard Deviation of the Mean being smaller, all the data points will tend to be very near the Mean. The Median with a larger Standard Deviation will tend to have data points spread out over a large range of values.
Since the Mean has the smaller of the Standard Deviations, it has the least variability
6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of interest? Explain your reasoning.
since the mean is a parameter of interest, it is a better estimate.
The mean seems to be a better estimate because the data points are closer to the mean and the standard deviation is less than the medians.
7. Give and interpret the 95% confidence interval for the hours of sleep a student gets.
Based on the information provided in questions 3, 4, and 5, we can determine whether the mean or median is a better estimate for the parameter of interest.
In question 3, we found that the mean for the mean column is 3.56, which is very close to the parameter of interest (3.5), but not equal to it. This means that the mean is a reasonable estimate, but not a perfect one.
In question 4, we found that the mean for the median column is 3.6, which is also close to the parameter of interest, but not as close as the mean from question 3.
In question 5, we compared the standard deviations of the mean and median columns. We found that the standard deviation of the mean column is 0.4762, while the standard deviation of the median column is 0.7539. Since the mean has a smaller standard deviation, it indicates that the data points tend to be very close to the mean. On the other hand, the median has a larger standard deviation, suggesting that the data points are spread out over a larger range of values.
Based on these findings, we can conclude that the mean is a better estimate for the parameter of interest. Although both the mean and median are close to the parameter, the mean is slightly closer and has less variability.