Instructions to obtain Chis quare test for Asseoieation output using Minitah:
* Store the IE'render in the eolumn Cl and Foot Symmetry in the rest of the eolumns of the Mnitah
* Choose Stat Tahles 4,: Chi-square test for Assoieaition.
Stepl:The null and alternative hypotheses are:
He : There is ne- relatie-nship between a person's ability in Mathemati es and their interest in statisties.
H1 :There is relati e-nship between a person's ability in Mathematies and their interest in stati s
Stepl:We set up the null hypothesis that there is no difference in success rates between the
new therapy and standard therapy.
HE, : There is no difference in success rates between the new therapy and standard therapy.
HE1 :There is difference in success
Stepl:We setup the null hype-thesis that the age is related to gender in such alcehelrelated
He : The age is not related to gender in such alc-shel-related accidents.
H1 :The age is not related td gender in such alc-shel-related accidents.
(a) Step1: The null and alternative hypotheses are:
EU: The eurrentruarltet shares do not differ from theuse 6f1990.
H1 :The eurrent market shares differ fr-aru the-se 6f1990.
Step2: The level 6f signifieanee is er 2 0.05.
Stap3. ' The eapeeted frequeneie
[ojlnstruetions to obtain Chisquare test for Asseoieation output using Mnitab:
* Store the Age in the eolumn Cl and Hours of sleep is in the rest of the eolumns of the Minitah
* Choose Stat Tahles Chis quare test for Assoieaition.
* Choose Sum
Instructions to obtain Chis quare test for Asscoicati on output using Minitab:
* Store the Age in the column Cl and Type of Pizza in the rest of the columns of the Minitab
* Choose Stat Tables 4,: Chi-s quare test for Assoicaition.
* Choose Sum
[a] The correlation test is used te determine whether age was related to essam
grades for students in high seheel.
(is) We must use independent t-test to determine whether male er
female high sehe-e-l students seered hetter en exams.
lie) We must use ANDR
Step11The null and alternative hypotheses are:
He : Depresssic-n and Ella-11] in elderly men are statistically independent.
H:1 : Depresssic-n and Elk-11D in el cfw_13le men are statistically dependent.
The level c-f significance is cfw_l=C|.C|l.
Instrnetiens te ehtain Chis quare test fer i'l'isseeieati en eutput using Mnitah:
* Stere the Dressing in the eelunin Cl and speed ef advertisement
is in the rest ef the eelunins ef the Minitah werl-Lsheet.
* Cheese Stat Tahles 4,: Chisquare test fer Asse
[ajSteplee set up the null hypothesis that the hours of sleep on week nights are
dependent of age.
He :The delivery time performance is independent ofwhether computer assisted ordering is used.
H1The delivery time performance is dependent ofwhether comput
Step1:We set up the null hypothesis that the proportions that order eaeh type ofpizza are the
same among the age groups.
He : The proportions that order eaeh type ofpizza are the same among the age groups.
HE1 :The proportions that order eaeh type of pizz
(a) Stepl:We set up the null hypothesis the type of ear, nose, or throatirritation is
independent ofthe age group at a level of signifieanee equal to .05
H0 : The type of ear, nose, or throat irritation is dependent ofthe age group.
H1 :Tse type of ear, n
Instructions to obtain Chis quaro tost for Asscoicati on output using Minitalo:
* Storo tho Typo ofiriitation in tho column Cl and Ago is in tho rost of tho columns of tho Minitab
* Chooso Stat Tablos 4; Chisquaro tost for Assoicaition.
Stepl:The null and alternative hvpotheses are:
He : The tvvo variables family structure and Drug hi StOI'jF are not associated.
H1Thetho variables farnilv structure and Drug historyr are associated.
The level of signicance is 3:0.05.
The assumptions E- and C are EEESSEI'F for an; inferenees derived from this printout to be valid
in this ease.
The population varianees are equal and unknoum.
The population trarianees are equal and known.
These two also assumptions to make inferenees.
(e) Betweens= 2. 13 ands=1.34.
The area tn the left at: = 2. 13 is ebtained b; using eseel funetien,=nennsdist(a.
The areais 0.0146.
The area tn the right at: 21.34 is obtained by using esteel funeti en,=nennsdistcfw_s.
The areais [19099.
The between area
(a) The number of ears made offer powerloelcs as a standard feature is abinomial distribution with
parameters Pi = 7", and: = [3.21
= l 0.2?
Let? be the random variable fellows a binomial distribution.
The pmf of binomial distributio
(a) The number oftrials is n =
The probability of resistant isp =
The probability of non-resi stant is g = 1:
= l 0.46
Let X be the random variable follows a binomial distribution.
The pmf of binomial distribution is
PIIX = Jr) = p [:le
(e) The mean of the number 6f passengers who show up is
EEK) = 92;:
= 126 [6.713)
E (if) = 93.23
(if) The sarianee 6f the number 6f passengers who show up is
VarUf) = Mpg
Therefore, the standard desiati sun of the num
The I;11'n:-l:ual:uilit5;r nf sueeess is; = 0.14.
The pfbabllltjf at failure is
The number of trails is as = 300.
LetX he the tandem sanahle fellows a binomial distnhuti en.
The pint of hinninial distribution is
In statistics, the 63 95 99.7" rule, also known as the three sigma rule or empirical rule,
states that nearlv all values lie within three standard deviations of the mean in a normal distribution.
About 68.23% of the values lie within one standard deviatio
The null and alternatiye hypotheses are:
[There is no signfieant differenee between the mean groups)
H1431: 2 i 3
[There is signfieant differenee between the mean groups)
The level of signifieanee is :1 23.05.
The eritieal yal
Fisher Pairwise Comparisons
Grouping Information Using the Fisher LSD Hethod and 95% Confidence
Factor N Mean Grouping
Treatment E 5 40.000 A
Treatment C 5 35.00 E
Treatment A 5 30.00 C
Means that do not share a letter are significantly different.