Average Tempreture of Texas in 2016 vs Crime Rate per Month in Houston Texas (TH
ID: 3324889 • Letter: A
Question
Average Tempreture of Texas in 2016 vs Crime Rate per Month in Houston Texas
(THIS IS WHAT I AM GETTING FROM THIS DATA PLEASE CHECK TO VERIFY)
Mean = 11307.67
Median = 10595.50
Minimum = 8956.00
Maximum = 20920.00
First Quartile 25th Percentile = 10298.25
Median Quartile 50th Percentile = 10595.5
Third Quartile 75th Percentile = 10811.5
Interquartile Range: = 513.25
Range = 11964.00
Standard Deviation = 3077.249283
Standard Variation = 9469463.152
1. • You will need a minimum of 30 observations from each group.
2. • Find the values of the descriptive measures (mean, median, quartiles, range, standard deviation, etc.)
3. • Using the Empirical rule, describe the data in the 95th and the 99th percentile. (Within each range, include both the percentage of data values predicted by the Empirical Rule and the actual percentage of data.)
4. • Construct a confidence interval estimate of the measure.
5. • Formulate a hypothesis (see chapters 8 and 9) based on your data from and analyze the data.
6 .• Complete an appropriate hypothesis test regarding the data. Be sure to formally state the null and alternate hypothesis, the decision rule and the conclusion.
7. • Analyze and compile your conclusions - summarize your conclusions.
(IF THERE IS CHANCE, PLEASE SHOW THE FORMULA FOR EXCEL)
Month Temprature Crime Rate January 54 10630.00 February 60 8956.00 March 67 10086.00 April 71 10149.00 May 76 10792.00 June 82 20920.00 July 85 10545.00 August 83 10870.00 September 82 10672.00 October 76 10561.00 November 68 10348.00 December 61 11163.00Explanation / Answer
I entered the data in excel and it is from cell A1 to C13
Here goes the calculations in excel along with the formulas:
Mean =
11307.67
=AVERAGE($C$2:$C$13)
Median =
10595.50
=MEDIAN($C$2:$C$13)
Minimum =
8956.00
=MIN(C2:C13)
Maximum =
20920.00
=MAX(C2:C13)
Q1 =
10298.25
=QUARTILE(C2:C13,1)
Q2 =
10595.5
=QUARTILE(C2:C13,2)
Q3 =
10811.5
=QUARTILE(C2:C13,3)
Interquartile Range
256.625
=(QUARTILE(C2:C13,3) - QUARTILE(C2:C13,1))/2
Range
11964.00
=MAX(C2:C13)-MIN(C2:C13)
Standard Deviation
3077.249283
=STDEV(C2:C13)
Variance
9469463.152
=VAR.S(C2:C13)
I see that there is nothing called Standard Variation in Statistics. It should be Variance. Also you have computed everything correctly but the interquartile range is not correct. you have calculated quartile range. the word 'inter' means it has to be divided by 2.
For the rest of the questions please provide a data set.
Month Temprature Crime Rate January 54 10630.00 February 60 8956.00 March 67 10086.00 April 71 10149.00 May 76 10792.00 June 82 20920.00 July 85 10545.00 August 83 10870.00 September 82 10672.00 October 76 10561.00 November 68 10348.00 December 61 11163.00Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.