can someone help with these problems

Question

An Arborist is interested in determining the mean diameter of mature white oak trees. The following data represent the diameter (in centimeters) 1 meter above the bases for a random sample of seven mature white oak trees in a forest preserve. 

64.0 33.4 45.8 56.0 51.5 29.2 63.7

Treat the sample as a simple random sample of all white oak trees in the forest preserve. 

Part 1

1. Find the sample mean and the sample standard deviation. 

mean = 49.08571; sd = 13.7957

1. What is the point estimate of the population mean diameter? Specify its value and statistical notation for this estimate. 
2. Find the standard error of this estimate. Explain what this number means. 
3. What is the value of t-multiplier tα/2,n−1 for a confidence interval? Round your answer to 3 decimal places. 
4. Find the margin of error for a 95% confidence interval. Round your margin of error to the nearest two decimal places. 

Part 2 

1. Find the 95% confidence interval of the mean diameter of mature white oak trees. Interpret the interval in context.
2. Can we conclude that the population mean diameter, 1 meter above the base, of mature oak trees in the forest preserve children is greater than 30 centimeters? 
3. Find the t-value such that the area in the right tail is 0.10 with 25 degrees of freedom.
4. Find the t-value such that the left of the t-value is 0.01 with 18 degrees of freedom.

Part 3

1. State assumptions of the confidence interval for the population mean diameter of mature white oak trees. Determine whether each assumption is met based on the Problem 3 description. Explain. Hint: (Draw a Q-Q plot to verify that the data come from a population that is normal, a boxplot to verify that there are no outliers, and a histogram. Submit all tree plots. 
2. (Multiple Choice Question) The histogram from the previous question is: 

i. A population distribution

ii. A data distribution (sample distribution) 

iii. A sampling distribution

Choose one and briefly explain. 

1. Use the statistical software R to construct a 95% confidence interval for the population mean diameter of a mature oak tree 1 meter above the bases using the sampling data given. (HINT: Use the t.test () command.) 
2. The following interprets the 95% confidence interval for the population mean diameter of a mature oak tree 1 meter above the base using the sample data from Problem 3 parts 2) and 3) (36.33 cm, 61.85 cm). What of the following is correct? 

i. In the entire population, 95% of the diameters of mature oak trees 1 meter above the base falls within 36.33 cm and 61. 85 cm. 

ii. We are 95% confident that the sample means of diameters, 1 meter above the base, of mature white oak trees in the forest preserve is between 36.33 cm and 61.85 cm.

iii. If we repeatedly sample 7 white oak trees randomly and construct 95% confidence intervals, then in the long run, 95% of those intervals contain the population mean diameter, 1 meter above the base, of mature white oak trees. 

iv. There is a 95% probability population mean diameter, 1 meter above the base, of mature white oak trees is between 36.33 cm and 61.85 cm

Answer 1

Given data are:

Sample size: n = 7

Sample data: [64.0 33.4 45.8 56.0 51.5 29.2 63.7]

Mean = 49.08571

Sample standard deviation = 13.7957

Analyzing the given list:

Ordered list: [29.2, 33.4, 45.8, 51.5, 56.0, 63.7, 64.0] 

Min: 29.2

Q1: 33.4

Q2: 51.5 (Center)

Q3: 63.7

Max: 64

50% data between Q1 and Q3

Part 1:

1.     What is the point estimate of the population mean diameter? Specify its value and statistical notation for this estimate. 

Answer:

Point estimate of the population mean = 49.086 cm

A single estimated value is a point estimate of mean

2. Find the standard error of this estimate. Explain what this number means. 

Answer:

Standard error = standard deviation of sample / sqrt( n)

                         = 13.7957 / sqrt(7) = 5.21428448

Standard error of sample expressed the variability of data around center of sample. As the size n increases, the standard error reduces to the population error.

3. What is the value of t-multiplier tα/2,n−1 for a confidence interval? Round your answer to 3 decimal places. 

Answer:

Depending of the confidence level, 0.01, 0.05, 0.1 and degree of freedom 7-1 =6, the Student’s t-distribution gives: (https://www.statdistributions.com/t/)

p-value (significance level) of 0.01: t-value (t*) = 3.143 (right tail)

p-value (significance level) of 0.05: t-value (t*) = 1.943 (right tail)

p-value (significance level) of 0.10: t-value (t*) = 1.440 (right tail)

4.  Find the margin of error for a 95% confidence interval. Round your margin of error to the nearest two decimal places. 

Answer:

At 95% confidence interval, t* = 1.943 and standard error = = 5.21428448,

Therefore,

margin of error (m) = t* (s /sqrt(n)) = 1.943 * 5.21428448

                              = 10.131

Part 2:

1.  Find the 95% confidence interval of the mean diameter of mature white oak trees. Interpret the interval in context.

Answer:

mean = 49.08571

Margin of error (m) = 10.131 cm

Interval of diameters at 95% confidence =

[49.08571 - 10.13135474,  49.08571 + 10.13135474]

= [38.95, 59.21]

At 95% confidence interval, the diameter of trees varies from the mean by 10.12 cm, up or down.

2. Can we conclude that the population mean diameter, 1 meter above the base, of mature oak trees in the forest preserve children is greater than 30 centimeters? 

Answer:

The least margin of error (m) at significance level 0.10 is  

t* (s /sqrt(n)) = 1.44 * 5.21428448

                     = 7.509 cm

While the greatest margin of error (m) at significance level 0.01 is  

t* (s /sqrt(n)) = 3.143 * 5.21428448

                    = 16.388 cm

Thus, a 30 cm diameter is guaranteed to be included at all confidence intervals.

3. Find the t-value such that the area in the right tail is 0.10 with 25 degrees of freedom.

Answer:

t* = 1.316 (from Student's t-distribution, https://www.statdistributions.com/t/)

4. Find the t-value such that the left of the t-value is 0.01 with 18 degrees of freedom.

nswer:

t* = -2.553