Question

Employees of a certain company took a mean of 6.8 vacation days in 2016. The CEO of the company believes that in 2017 the average number of sick days was less than 6.8. A sample of 52 employees took a mean of 6.6 days in 2017, σ is known to be equal to 1.7. At α = 0.02, test the CEO’s claim.
a 1-6) Give the hypotheses for H₀ (a1, a2 and a3) and H₁ (a4, a5 and a6)
H₀
a1) µ or p

µpClick for List  
a2) =, ≥, or ≤

=≥≤Click for List  
a3)

H₁
a4) µ or p

µpClick for List  
a5) ≠, >, or <

≠&gt;&lt;Click for List
a6)

b) Calculate the test statistic. z = _______ (Round your answer to 3 decimals.)

c 1-3) Formulate the decision rule for the critical value approach.
Reject H₀ if
c1) z or p

zpClick for List  
c2) < or >

&lt;&gt;Click for List  
c3)

d 1) Make a decision.
d1) Reject Ho or Do not reject Ho

Reject HoDo not reject HoClick for List  

e1-2) Give your conclusion. At α = .___, there (is/is not) enough evidence to conclude that in 2017 the average number of sick days was less than 6.8.

e1)

   
e2) is or is not

isis notClick for List  

Answer 1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 6.8 or mu >= 6.8
Alternative Hypothesis, Ha: μ < 6.8

b)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (6.6 - 6.8)/(1.7/sqrt(52))
z = -0.848

c)

Rejection Region
This is left tailed test, for α = 0.02
Critical value of z is -2.054.
Hence reject H0 if z < -2.054

d)

Do not reject Ho

e)

At α = .0.02, there (is not) enough evidence to conclude that in 2017 the average number of sick days was less than 6.8.