H0: = < > 0 vs. HA: > < 0 n = _______ = _______

• Testing a Hypothesis About a Mean;• Process Performance Measure is Approximately Normally

Distributed or We Have a “small” Sample;• We Do Not Know Which Must be Estimated by S.• Therefore this is a “t-test” - Use Student’s T Distribution.

Calculate tcalc = (X - 0)/(s/ n ) DR: ( in HA) Reject H0 in favor of HA if tcalc < -t/2, n-1 or if

tcalc > +t/2,n-1. Otherwise, FTR H0. DR: (> in HA) Reject H0 in favor of HA iff tcalc > +t n-1 .

Otherwise, FTR H0. DR: (< in HA) Reject H0 in favor of HA iff tcalc < -t, n-1

Otherwise, FTR H0.

t-test ,To test one population mean

Degrees of Freedom

# in sample - number of parameters that must be estimated before test statistic can be computed.

For a single sample t-test, we must first estimate the mean before we can estimate the standard deviation.Once the mean is estimated, n-1 of the values are left since we know that the nth value is equal to

1

1

n

iixxn

Example

First People’s Bank of Central City would like to improve their loan application process. In particular currently the amount of time required to process loan applications is approximately normally distributed with a mean of 18 days.

Measures intended to simplify and speed the process have been identified and implemented. Were they effective? Test the appropriate hypothesis at the = .05 level of significance if a sample of 25 applications submitted after the measures were implemented gave an average processing time of 15.2 days and a standard deviation of 2.0 days.

H0: > 18 HA: < 18 n = 25 = .05 (i) testing a mean (ii) normal distribution (iii)

is unknown and must be estimated so that this is a t-test

DR: Reject H0 in favor of HA iff tcalc < -1.711.

Otherwise, FTR H0 tcalc = (X - 0)/(s / √n) = (15.2 - 18)/ (2/ √ 25 )

= -2.8/.4 = -7.00 X Reject H0 in favor of HA.

Example

The weight of a canned food product is specified as 500 grm. For a sample of 8 cans the weight were observed as 480, 475, 510, 500, 505, 495, 504 and 515 grm. Test at 5% level of significance, whether on an average the weight is as per specification.Table value = 2.365

H0: = < > 2 vs. HA: > < 2 n = _______ = _______

• Testing a Hypothesis About a Mean;• Process Performance Measure is Approximately Normally

Distributed or We Have a “small” Samples;• We Do Not Know Which Must be Estimated by S.• Therefore this is a “t-test” - Use Student’s T Distribution.

Calculate (x1 - x2) - (1 - 2 ) t = -------------------------

s* ( 1/n1 + 1/n2 )

with d.f. = n1 + n2 - 2. In this expression, s* is the pooled standard deviation, given by (n1 – 1)s1

2 + (n2 – 1)s22

s2

= --------------------------------- n1 + n2 - 2

t-test ,To test two population mean

DR: ( in HA) Reject H0 in favor of HA if tcalc < -t/2, n-1 or if tcalc > +t/2,n-1. Otherwise, FTR H0.

DR: (> in HA) Reject H0 in favor of HA iff tcalc > +t n-1 . Otherwise, FTR H0.

DR: (< in HA) Reject H0 in favor of HA iff tcalc < -t, n-1 Otherwise, FTR H0.

T- Test for Independent Samples - Two Cities

n1 = 20 x1 = 30 s1 = 22n2 = 10 x2 = 35 s2 = 25

The degrees of freedom for the t‑statistic are d.f. = 28

Critical value of t with 28 d.f for a tail probability of .025 is 2.05.

Decision rule : “Reject H0 if t -2.05 or if t 2.05." The pooled standard deviation is

s* = 529 (approximately) = 23The test statistic is t = -.56Since t is neither less than -2.05 nor greater than 2.05, we cannot reject H0

The sample evidence is not strong enough to conclude that the two cities differ in terms of levels of exercising activity of their residents.

Cross-Tabulations: Chi-square Test

Technique used for determining whether there is a statistically significant relationship between two categorical (nominal or ordinal) variables

Telecommunications Company

Marketing manager of a telecommunications company is reviewing the results of a study of potential users of a new cell phone

Random sample of 200 respondentsA cross-tabulation of data on whether target consumers would buy the phone (Yes or No) and whether the cell phone had Bluetooth wireless technology (Yes or No)

QuestionCan the marketing manager infer that an association exists between Bluetooth technology and buying the cell phone?

Two-Way Tabulation of Bluetooth Technology and Whether Customers Would Buy Cell Phone

Cross Tabulations -Hypotheses

H0: There is no association between wireless technology and buying the cell phone (the two variables are independent of each other).

Ha: There is some association between the Bluetooth feature and buying the cell phone (the two variables are not independent of each other).

Conducting the Test

Test involves comparing the actual, or observed, cell frequencies in the cross-tabulation with a corresponding set of expected cell frequencies (Eij)

Expected Values ninj

Eij = ----- n

Where ni and nj are the marginal frequencies, that is, the total number of sample units in category i of the row variable and category j of the column variable, respectively

Computing Expected Values

The expected frequency for the first-row, first-column cell is given by

100 100 E11 = ------------ = 50

200

Observed and Expected Cell Frequencies

Chi-square Test Statistic

r c (Oij - Eij)2

2 = -----------------

i=1 j=1 Eij

= 72.00

Where r and c are the number of rows and columns, respectively, in the contingency table. The number of degrees of freedom associated with this chi‑square statistic are given by the product (r - 1)(c - 1).

Chi-square Test Statistic in a Contingency Test

For d.f. = 1, Assuming =.05, from Appendix 2, the critical chi‑square value (2

c) = 3.84.

Decision rule is: “Reject H0 if 2 3.84.”

Computed 2 = 72.00Since the computed Chi-square value is greater than the critical value of 3.84, reject H0. The apparent relationship between “Bluetooth technology"and "would buy the cellular phone" revealed by the sample data is unlikely to have occurred because of chance

In a management institute, the A+, A and B grades allocated to students in there final examination, were as follows. Using 5% level of significance, determine whether the grading scale is independent of the specialization.Table value = 9.488

SpecializationGrade Finance Marketing

Operations  A+ 20 15 05 A 25 20 15

B 10 08 07