# HI6007 Statistics And Research Methods For Business Decision Making – Individual Assignment Help

## Question 1

(a). Arrange the data of 20 student’s result

 Student number Results 1 42 2 53 3 54 4 61 5 61 6 61 7 62 8 63 9 64 10 66 11 67 12 67 13 68 14 69 15 71 16 71 17 76 18 78 19 81 20 83 1318

• Compute Mean, Median and Mode

Formula of calculating Mean:

Σ xi=1318;    n=20

131820= 65.9

Formula of calculating Median

Median = N/2 +1 = (20/2)+1 = 10+1 =11th Item = 67

• Compute 1st and 3rd Quartile

1st Quartile =  14 N+1 = 20+1/4 = 5th Item  = 61

3rd Quartile =34N+1=3420+1=634=15 th item=71

• Compute and Intercept 90th Percentile

Formula = 90% of N =  0.90*20 = 18th Item = 78

(b) Inferential Statistics:

Inferential statistics is the process of obtaining data from the sample available, in order to make estimates  and there after applying appropriate test, i.e. z-score, t-test, in order to test Hypothesis regarding the population characteristics.

## Question 2

(i) Prepare Joint Probability Table

 Applied for More than 1 University Age Group Yes No 23 and Under 207 207/808*100=25.62% 201 201/1210*100=16.61% 24-26 299 299/808*100=37.0% 379 379/1210*100=31.32% 27-30 185 185/808*100 = 22.90% 268 268/1210*100=22.15% 31-35 66 66/808*100=8.17% 193 193/1210*100= 15.95% 36 and over 51 51/808*100=6.31% 169 169/1210*100= 13.97% Total / Joint Probability(%) 808 100% 1210 100%

(ii) Given that a student applied to more than 1 university, what is the probability that the student is 24-26 years old.

Probability of student is 24-26 years old = 299/808 =37.00%

• Is the number of universities applied to independent of student age?Explain

Yes, The case described in the question student age is an independent factor . this indicates that at any age  student can enrol in one university and more than one university. Enrolment in one university or more than one  is depending variable.

(b)

 x f(x) 10 0.05 20 0.1 30 0.1 40 0.2 50 0.35 60 0.2 Total

X represent number of new clients for counselling cases in the year 2021.

Formula of calculating Expected value =  𝐸(𝑥) = 𝜇 = ∑𝑥 𝑓(𝑥)

 x f(x) (𝑥 − 𝜇) 2 (𝑥 − 𝜇) 2*f(x) 10 0.05 -33 1089 54.45 20 0.1 -23 529 52.9 30 0.1 -13 169 16.9 40 0.2 -3 9 1.8 50 0.35 7 49 17.15 60 0.2 17 289 57.8 Total 201

Expected Value= (10*0.05+20*0.1+30*0.1+40*0.2+50*0.35+60*0.2) =43

Formula of Variance of a discrete random variable 𝑉𝑎𝑟(𝑥) = ∑(𝑥𝜇) 2 𝑓(𝑥

Variance = 201 (calculation shows in table)

## Question 3

1. Formulate Hypothesis :

Problem statement: Population annual expenditure on prescription drugs per person is lower in the Midwest than the Northeast.

The above problem can be tested on one tail test from left  tail as it requires testing of lower limit.

Hypothesis Statement:

Ho: µ ≤ \$838 or

Ho: µ = \$838

Ha: µ > \$838

One tail test

• Suitable test Statistics

One (Left) tail test

Formula:

• Calculate value of relevant test statistics and P- value

Sample Mean (x) = \$745

Null Hypothesis Mean  = \$838

SD = 300

Sample size = 60

Applying Formula (745-838)/300/sqrt(n)

Z=  -93/38.75 = -2.40

From the table given of Z score , at significance level of 0.05 ,

P value = 0.0071

• Based on the p value in part (III), at 99% confidence level, decide the decision criteria.

At 1% significance level  the critical value  is 2.326, the Z score is -2.4 which is lower than the critical value , so the null hypothesis shall be rejected  and Alternative hypothesis shall be accepted.

• Make  the conclusion Based on the analysis.

The calculations above has indicated that the bull hypothesis is rejected as the expenditure of prescription drugs per person is not lowest in Midwest than Northeast. Midwest has higher consumption of drugs as comparison of Northeast.

## Question 4

•  State the null and alternative hypothesis for single factor ANOVA to test for any significant difference in the mean price of gasoline for the three brands.

Hypothesis

H0 = µ1= µ2= µ3

H1 = µ1≠µ2≠ µ3

(ii) State the decision rule at 5% significance level.

Reject the H0 id t stat  > Z critical value, Other wise accept the null hypothesis

(iii) Calculate the test statistics

 A B C 3.77 3.83 3.78 3.72 3.83 3.87 3.87 3.85 3.89 3.76 3.77 3.79 3.83 3.84 3.87 3.85 3.84 3.87 3.93 4.04 3.99 3.79 3.78 3.79 3.78 3.84 3.79 3.81 3.84 3.86 Sample Mean 3.811 3.846 3.85 Varience 0.003349 0.004844 0.00382

ANOVA one- way test Formula

Formula F= MSTR / MSE

MSTR = 𝑆𝑆𝑇𝑅 / 𝑘 – 1

MSE = SSE /𝑛r – k

𝑥Ӗ= (3.81 + 3.84 + 3.85)/3 = 3.83

SSTR=  10(3.81- 3.83)+ 10(3.84-3.83)2 + 10(3.85-3.83)= 0.009

MSTR = 0.009/ (3-1) = 0.0045

P-value and critical value approaches

Value of test statistic

SSE = 9(0.003) +9(0.005) + 9(0.004) =0.108

MSE = 0.108/(30-3) = 0.004

F= 0.0045/0.004 =1.125

ANOVA Table

 Source of variation Sum of Squares Degrees of Freedom Mean Square F P- value Treatment 0.009 2 0.0045 1.125 0.044 Error 0.108 27 0.004 Total 0.117 29 0.0085

P- value calculation

Here Numerator df = 2; Denominator Df = 27 then the value of F at 0.01 = 5.49

Decision on the basis of test

The p-value < .05, So null hypothesis shall be rejected

Decision as per critical value approach

Based on an F distribution with 2 numerator d.f. and 27 denominator d.f., F.05 = 3.35.

Reject H0 if F > 3.35

Here F =1.125 > 3.35 which is evidence for rejection of null hypothesis.

(d) Based on the calculated test statistics decide whether any significant difference in the mean price of gasoline for three bands.

Yes the calculation indicates that null hypothesis is rejected which means there is significance difference  in the mean prices of gasoline for all the three bands.

## Question 5

• Complete the missing entries from A to H in this output

A= R Square = SSR/SST = 35250755.68/ 42699148.82 = 0.82

B= Observation = 50 (provided in question)

C= residual = Total- Regression = 49-2 = 47

D= 42699148.82-7448393.14 = 35250755.68

E= SSRegression / dfreg. = 35250755.68/2 = 17625377.8

F= SSR /(50-3) = 7448393.148/47 = 158476.45

G= 17625377.8/158476.45 = 111.217647

H= Coefficient of income /Standard error of income = 8.36

• Estimate the annual credit card charges for a three-person household with an annual income of \$40,000

To estimate  charges of credit card , intercept, household value and size has been considered from the ANOVA table.

The calculation indicates the annual credit card charges will be \$3700 for three person household.

• Did the estimated regression equation provide a good fit to the data? Explain

No,  The regression equation did not6 provide a good fit to data, as there is high variability between data of x- values and y- values.

## Question 6

1. Using linear trend equation forecast the sales of face masks for October 2020
 Month Sales (\$) 1 17000 2 18000 3 19500 4 22000 5 21000 6 23000

Linear Trend Equation = Y= Mx+B

M= Y2-Y1X1-X2 = 23000-21000/6-5 = 2000/1 = 2000

Y = Mx+B

Y= ?

X=1

B= 23000

Y = 2000*1+23000

Y = 2000+23000= 25000

• Sales forecast will be
 Sales Weight Weighted Sale July 22,000 0.2 4400 August 21,000 0.3 6300 September 23,000 0.5 11500 Total 22200
• So, the expected sale for the next month will be \$22200