--%>

Compute two sample standard deviations

Consider the following data for two independent random samples taken from two normal populations.

Sample 1 14 26 20 16 14 18

Sample 2 18 16 8 12 16 14

a) Compute the two sample means and the two sample standard deviations.

b) What is the point estimate of the difference between the two population means?

c) Assuming α = .10, conduct p-value based and critical-value based hypothesis tests for the equality

of means of the two populations.

d) What is the 90% confidence interval estimate of the difference between the two population means?

How do the results compare in all the three approaches to hypothesis testing?

 

E

Expert

Verified

Mean sample 1 = X1-bar = (14+26+20+16+14+18)/6 = 18

Mean sample 2 = X2-bar = (18+16+8+12+16+14)/6 = 14

Sample 1 SD = SD1

X1

X1-X1-bar

(X1-X1-bar)2

14

-4

16

26

8

64

20

2

4

16

-2

4

14

-4

16

18

0

0

Sum of (X1-X1-bar)2 = 104

S12 = 104/6-1

        = 20.8

SD1 =  = 4.56

Sample 2 SD = SD2

X2

X1-X1-bar

(X1-X1-bar)2

18

4

16

16

2

4

8

-6

36

12

-2

4

16

2

4

14

0

0

 

Sum of (X2-X2-bar)2 = 64

S22 = 64/6-1

        = 12.8

SD2 =  = 3.58

(b)

Point estimation of difference b/w two means = 18 - 14 = 4

(c)

t-test will be applied because sample size is small.

Hypothesis Formation

Null Hypothesis H0:    µ1 - µ2 = 0

Alternative Hypothesis H1:    µ1 - µ2 ≠ 0

t Statistic

t-statistic = (X1-bar  - x2-bar)/Sp

Where SP =

                  = 2.016

Critical value

Critical value of t with df=10 at 0.1 significance level = 1.812

Critical Region

Reject null hypothesis in favor of alternative if t is greater than t critical value of 1.812 or less than -1.812.

Computation

t-statistic = (18 - 14)/2.016

   = 5.95

Decision

Null hypothesis is rejected in favor of alternative as Z value is greater than Z critical value.

(d)

90% CI of difference between means = (18-14) - 1.812*2.016

                                                                    = 4 - 1.22 < µ < 4 + 1.22

                                                                    = 2.78< µ< 5.22

   Related Questions in Basic Statistics

  • Q : Point of estimate standing data se to

    standing data se to develop a point of estimate

  • Q : Hypothesis homework A sample of 9 days

    A sample of 9 days over the past six months showed that a clinic treated the following numbers of patients: 24, 26, 21, 17, 16, 23, 27, 18, and 25. If the number of patients seen per day is normally distributed, would an analysis of these sample data provide evidence that the variance in the numbe

  • Q : Problem on queuing diagram Draw a 

    Draw a queuing diagram for the systems below and describe them using Kendall’s notation: A) Single CPU system <

  • Q : Define Operational Analysis

    Operational Analysis: • Analysis method based on the measurement of the operational characteristics of the system.

    Q : Calculate the p- value Medical tests

    Medical tests were conducted to learn about drug-resistant tuberculosis. Of 284 cases tested in New Jersey, 18 were found to be drug- resistant. Of 536 cases tested in Texas, 10 were found to be drugresistant. Do these data indicate that New Jersey has a statisti

  • Q : Correlation analysis and the regression

    1).  When you take out a mortgage, there are many different kinds of costs.  Usually the two largest are the interest rate (annual percentage that determines the size of your monthly payment) and the loan fee (a one-time percentage charged to you at the time

  • Q : Cumulative Frequency and Relative

    Explain differences between Cumulative Frequency and Relative Frequency?

  • Q : What is Inter-arrival times

    Inter-arrival times:A) Requests arrive randomly, often separated by small time intervals with few long separations among themB) The time until the next arrival is independent of when the last arrival occurredC) Coro

  • Q : Report on Simple Random Sampling with

    One of my friend has a problem on simple random sampling. Can someone provide a complete Report on Simple Random Sampling with or without replacement?

  • Q : Explain Service times Service times: A)

    Service times:A) In most cases, servicing a request takes a “short” time, but in a few occasions requests take much longer.B) The probability of completing a service request by time t, is independent of how much tim