Question

(a) Show that the sample variance s 2 = [Pn i=1(xi − x¯) 2 ]/(n − 1) can also be expressed as s 2 = [Pn i=1 x 2 i − ( Pn i=1 xi) 2 n ]/(n − 1). At a medical center, a sample of 36 days showed the following number of cardiograms done each day.

25 31 20 32 20 24 43 22 57 23 35 22 43 26 56 21 19 29 36 32 33 32 44 32 52 44 51 45 47 20 31 27 37 30 18 28

(b) (1 point) Find the sample mean ¯x and the sample variance s 2 x .

(c) (2 points) Construct a stem and leaf plot for the data and find the sample median.

(d) (3 points) Construct a 86% confidence interval for the population µ.

(e) (4 points) A researcher wishes to test the claim that the average number of cardiograms done each day is equal to or greater than 33. Is there evidence to support the claim at α = 0.05? Find the p-value.

(f) (1 point) Let x1, x2, · · ·, x36 be the data of 36 days of cardiograms above; let a and b be any nonzero constants. If y1 = a x1 +b, y2 = a x2 +b, · · ·, y36 = a x36 +b, and let ¯y and s 2 y be the sample mean and the sample variance of the yi ’s, respectively. What is the relationship between ¯x and ¯y? What is the relationship between s 2 x and s 2 y ?

(g) (5 points) Show that s 2 x is an unbiased estimator of the population mean σ 2 .

Answer 1

a). Sample variance s2 =   =  

=  

=  

b). Sample mean = = = 32.97 ....here (n=36)

Sample variance = s2 =   = 123.056

c). Stem and Leaf Plot:

Sample Median = 31.5

d). Population = sample mean = 32.97

Confidence interval formula:

X ± Z	s
√(n)

Where:

• X is the mean
• Z is the Z-value from the table ( Z-value for 86% confidence is approx: 1.48)
• s is the standard deviation
• n is the number of observations

The required confidence interval is: 32.97 ± 2.74
(30.23 to 35.71)

e). Hypothesis (Ho): 33 , α = 0.05

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation. Here x=33

z=-0.02 ,p-value = .492022

Here, We cannot reject the null hypothesis. The z score of -0.02 is within the nonrejection area. The critical value (the cutoff point) is 1.645. In right-tail hypothesis testing, any z score less than the critical value cannot be rejected. Since -0.02 is less than 1.645, we cannot reject the null hypothesis