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APPLIED STATISTICS QUIZ 20 POINTS TUESDAY, Jan. 2 2019 Q#1 A personal consultant was hired to study the influence of sick-pay benefits on absenteeism. She randomly selected samples of hourly employees who do not get paid when out sick and salaried employees who receive sick pay. Using the following data on the number of days absent during a 1- year period. HOURLY 10 12 A. State the Null hypothesis and the Research hypothesis B. Test the null hypothesis at alpha 0S that hourly and salaried employees do not differ with respect to absenteeism C. Show all the six steps of the t tests. D. Write the conclusion based on your findings. Q2. Even in the darkest areas of human behavior, we tend to imitate our heroes. A. sociologist studied the incidence of suicide in five randomly selected communities of moderate size both before and after publicity was given to the suicide of a famous singer Using the following data on the about suicide had no effect on suicide. stzng data on the number of suicides, test the null hypothesis that Before After Community 10 A. State the Null hypothesis and the Research hypothesis B. Test the null hypothesis at alpha 05 that publicity about suicide had no effect on suicide C. Show all the six steps of the t tests E. Write the conclusion based on your findings.

Answer 1

2 sample t test :

Claim : Hourly and salaried employees do not differ with respect to absenteeism.

A) null and alternative hypothesis :

H₀: µ₁ = µ₂    vs H_a: µ₁ ≠ µ₂   

We can find the sample mean and sample standard deviation of the both data set using excel function =AVERAGE( ) and =STDEV.S() respectively.

Salaried Hourly 0 4 4 2 10 12 4 10 0 114 12 13 AVERAGE(D2:D11) 5.2 -STDEV.S(D2:D11 3.7653 Mean AVERAGE(A2:A11) 3.1 STDEV.S(A2:A11) 2.2336 Mean SD SD

B) So we have $\bar{x}$₁ = 3.1 , ; S₁ = 2.2336 , n₁ = 10 and $\bar{x}$₂   = 5.2, S₂= 3.7653, n₂ = 10

Pooled estimate of standard deviaion :

s = $\sqrt{\frac{(n_{1}-1)*s_{1}^{2}+(n_{2}-1)*s_{2}^{2}}{n_{1}+n_{2}-2}}$

s = $\sqrt{\frac{(9*2.2336^2)+(9*3.7653^2)}{10+10-2}}$

s = 3.0957

C) Test statistic :

t =  $\frac{\bar{x}_1 -\bar{x}_2 }{{s}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$

t =  $\frac{3.1 - 5.2}{3.0957*\sqrt{\frac{1}{10}+\frac{1}{10}}}$

t = -1.52

Critical value :

We are given α =0.05

Test statistic follows t distribution with degrees of freedom (df) = n₁+ n₂ - 2 = 10+10-2 = 18

We can find critical value using excel function =TINV( α , d.f )

=TINV( 0.05 , 18 ) = 2.1009

Critical value = 2.1009

Critical region: Reject H₀, if | t | ≥ 2.1009 Or fail to reject H0 , if |t| < 2.1009

|t |is absolute value of t statistic.

Decision : we have t = -1.52 , so |t| = 1.52

As |t| < 2.1009, we fail to reject H₀

D) Conclusion : We do not have significant evidence that Hourly and salaried employees differ with respect to absenteeism.