north south university tutorial 6 - university of...

NORTH SOUTH UNIVERSITYTUTORIAL 6

AHMED HOSSAIN,PhD

Biostatistics I

AHMED HOSSAIN,PhD - Biostatistics I 1

Hypothesis TestingSTEPS IN HYPOTHESIS TESTING

1 State the hypotheses (both null and alternative).2 Specify the significance level (↵).3 Draw sample of size n, compute the test statistic. For example, use normal (Z )

when � is known and t when you have estimated � with s.4 Determine p-value5 Compare p-value to the significance level ↵ and decide whether or not to reject

H0

6 State conclusions in terms of subject matter.


Tests of the Equality of Two MeansWHEN POPULATION VARIANCES ARE EQUAL

HYPOTHESIS TESTING H0 : µX � µY = 0 versus H1 : µX � µY 6= 0Test statistic, T = (X�Y )

SE(X�Y )where,

SE(X � Y ) = Sp

s1n1

+1n2

S2p =

(n1 � 1)S2X + (n2 � 1)S2

Yn1 + n2 � 2

The T is following t distribution with n1 + n2 � 2 degrees of freedom.


Example: Tests of the Equality of Two MeansWHEN POPULATION VARIANCES ARE EQUAL

EXAMPLE A psychologist was interested in exploring whether or not male andfemale college students have different driving behaviors. Sheconducted a survey of a random n1 = 34 male college students anda random n2 = 29 female college students. From the survey shefound x = 105.5 and y = 90.9 for male and female, respectively.Again, she found the standard deviations as 20.1 and 12.2 for maleand female, respectively. Is there sufficient evidence at the ↵ = 0.05level to conclude that the mean fastest speed driven by male collegestudents differs from the mean fastest speed driven by female collegestudents?

SOLUTION H0 : µM � µF = 0 versus H1 : µM � µF 6= 0.Using the test statistic, t = 3.42 where sp = 16.9.The critical value approach tells us to reject the null hypothesis infavor of the alternative hypothesis if:

|t | � t↵/2,n1+n2�2 = t0.025,61 = 1.9996

We reject the null hypothesis because the test statistic (t = 3.42) fallsin the rejection region.


Tests of the Equality of Two MeansWHEN POPULATION VARIANCES ARE KNOWN

HYPOTHESIS TESTING H0 : µ1 � µ2 = 0 versus H1 : µ1 � µ2 6= 0Test statistic, Z = (X�Y )

SE(X�Y )where,

SE(X � Y ) =

s�2

1n1

+�2

2n2

The Z is following standard normal distribution. Again, when thepopulation variances are not known and unequal then the teststatistic is

T =(X � Y )

SE(X � Y )

where, SE(X � Y ) =

rs2

n1+

s22

n2.

The T is following t distribution with n1 + n2 � 2 degrees of freedom.


Example: CI and Tests of the Equality of Two MeansWHEN POPULATION VARIANCES ARE KNOWN

EXAMPLE Suppose you are interested to examine the effectiveness of asmoking cessation program for pregnant women. The mean numberof cigarettes smoked daily at the close of the program by the 328women who completed the program was 4.3 with a standarddeviation of 5.22. Among 64 women who did not complete theprogram, the mean number of cigarettes smoked per day at the closeof the program was 13 with a standard deviation of 8.97. Construct a99% confidence interval for the difference between the means of thepopulations from which the samples may be presumed to have beenselected. Also test the hypothesis for equal means (Not Solved here).

SOLUTION The estimated standard error is

SE(X1 � X2) =

s5.222

328+

8.972

64= 1.1577

Therefore the 99% confidence interval for the difference betweenpopulation means is

�8.7 ± 2.58(1.1577)

�11.7,�5.7


ExercisesCONFIDENCE INTERVAL AND HYPOTHESIS TESTING

Q1 In a study of patient flow through the offices of general practitioners, itwas found that a sample of 35 patients were 17.2 minutes late forappointments, on the average. Previous research had shown thestandard deviation to be about 8 minutes. What is the 90%confidence interval for the true mean of time late for appointments?Test whether the population mean of the late for appointment is 36.

Q2 A sample of 25 freshman nursing students made a mean score of 77on a test designed to measure attitude toward the dying patient. Thesample standard deviation was 10. Do these data provide sufficientevidence to indicate, at the .05 level of significance, that thepopulation mean is less than 80? What assumptions are necessary?

Q3 A case-control study was carried out on 50 patients at a Hospital.The outcome was the number of inpatient treatment days forpsychiatric disorder during the year following the end of the program.Among 18 subjects with schizophrenia, the mean number oftreatment days was 4.7 with a standard deviation of 9.3. For 10subjects with bipolar disorder, the mean number of psychiatricdisorder treatment days was 8.8 with a standard deviation of 11.5.Construct a 95% CI for the difference between the means of thepopulations represented by these two samples. Also test whether thepopulation means are equal.


Population ProportionCONFIDENCE INTERVAL AND HYPOTHESIS TESTING

PROPORTION p = x/n

STANDARD ERROR OF PROPORTION SE(p) =q

p(1�p)n

CONFIDENCE INTERVAL 100(1 � ↵) percent confidence interval for p is given by

p ± z1�↵/2SE(p)

HYPOTHESIS TESTING H0 : p = p0 versus H1 : p 6= p0

z =p � p0qp0(1�p0)

n


Example: CI and TestsPOPULATION PROPORTION

EXAMPLE The Pew Internet and American Life Project (A-13) reported in 2003that 18 percent of Internet users have used it to search forinformation regarding experimental treatments or medicines. Thesample consisted of 1220 adult Internet users, and information wascollected from telephone interviews. Construct a 95 percentconfidence interval for the proportion of Internet users in the sampledpopulation who have searched for information on experimentaltreatments or medicines.

SOLUTION The estimated standard error is

SE(p) =

s.18(1 � .18)

1220= 0.0110

Therefore the 95% confidence interval for p is

0.18 ± 1.96(0.0110)

0.158, 0.202



EXAMPLE Suppose in a study you found that among 2428 boys ages 7 to 12years, 461 were overweight or obese. On the basis of this study, canwe conclude that more than 15 percent of the boys ages 7 to 12 inthe sampled population are obese or overweight? Let ↵ = .05.

SOLUTION H0 : p 0.15 versus H1 : p > 0.15 The estimated standard errorunder the null hypotheis is

SE(p) =

s.15(1 � .15)

2428= 0.007

Therefore test statistic

z = (0.19 � .15)/0.007 = 5.71

The p-value is < .001 and therefore the the null hypothesis can berejected at 5% significance level.


Test of HomogeneityCHI-SQUARE TEST


Chi-square Table


north south university tutorial 6 - university of...

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