Question

You will be performing an analysis on female heights, given of set of 30 heights that were randomly obtained. For this project, it is necessary to know that the average height for women is assumed to be 65 inches with a standard deviation of 3.5 inches. You will use these numbers in some of your calculations.

a. As you know, a random sample of 30 women’s heights was obtained. Describe the sampling distribution. Use a full sentence here to describe the sampling distribution. HINT: See the Chapter 8 Lecture Notes Key or the Chapter 8 video lecture if you are stuck. You will not use the raw data yet.

b. In StatCrunch or Excel, find and state the mean of the 30 women’s heights that were provided (round to two decimal places). Then (using the values from step 3), calculate the probability that a random sample of 30 women’s heights would result in a mean that was the value you found or more. Use this probability in a full sentence.

c. For this step, please work under the assumption that we do not know the population mean and standard deviation (and in fact, if we are running this test, it means that we are not sure of these values). Construct a 95% confidence interval for the average height of a female. Interpret this confidence interval. HINT: You can use the “with data” option

Height (in Inches)	Name
72.44	Emma
67.53	Olivia
66.71	Ava
62.02	Isabella
73.89	Sophia
65.95	Mia
65.83	Charlotte
64.15	Amelia
65.39	Evelyn
59.68	Abigail
64.24	Harper
66.60	Emily
65.40	Elizabeth
64.72	Avery
67.11	Sofia
61.97	Ella
62.83	Madison
67.20	Scarlett
66.62	Victoria
68.78	Aria
66.13	Grace
64.47	Chloe
66.64	Camila
62.39	Penelope
63.90	Riley
62.97	Layla
59.31	Lillian
66.14	Nora
67.54	Zoey
63.45	Mila

Answer 1

a) sampling distribution-

µx = 65

σx = σ/√n = 3.5/√30 = 0.6390

b) Sample Mean,    x̅ = ΣX/n =    65.4000

Z- score= (x̅ - µ )/SE = (   65.400   -   65   ) /    0.6390   =   0.626

P( x̅ > 65.4) = P(z>0.626) = 0.2657

the probability that a random sample of 30 women’s heights would result in a mean that was 65.4 or more = 0.2657

c)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   3.1125
Sample Size ,   n =    30
Sample Mean,    x̅ = ΣX/n =    65.4000

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   29          
't value='   tα/2=   2.045   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.1125   / √   30   =   0.5683
margin of error , E=t*SE =   2.0452   *   0.5683   =   1.1622
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    65.40   -   1.162244   =   64.2378
Interval Upper Limit = x̅ + E =    65.40   -   1.162244   =   66.5622
95%   confidence interval is (   64.24   < µ <   66.56   )