Question

6. Last year it was found that on average it took students 20 minutes to fill out the forms required for graduation. This year the department has changed the form and asked graduating student to report how much time it took them to complete the forms. Of the students 22 replied with their time, the average time that they reported was 18.5 minutes, and the sample standard deviation was 5.2. Can we conclude that the new forms take less time to complete than the older forms? Use a 0.1 significance level (i.e., p-value). (Assume that the reported times follow a Gaussian distribution)

Answer 1

Solution Q 6

Let X = Time taken (in minutes) to fill out the forms required for graduation.

Then, given X ~ N(µ, σ²) [Assume that the reported times follow a Gaussian distribution.]

Claim: The new forms take less time to complete than the older forms.

Hypotheses:

Null H₀: µ = µ₀ = 20   Vs Alternative H_A: µ < 20 [claim]

[Last year it was found that on average it took students 20 minutes to fill out the forms required for graduation and the claim.]

Test statistic:

t = (√n)(Xbar - µ₀)/s, where given

n = sample size = 22 ;

Xbar = sample average = 18.5 ;

s = sample standard deviation =5.5 .

So, t_cal = (√22(18.5 - 20)/5.5

= - 1.2792

Distribution, Level of Significance, α, Critical Value and p-value

Under H₀, t ~ t_{n – 1}

Critical value = lower α% point of t_{n – 1} = lower 10% point of t₂₁ = - 1.3231 [given,’Use a 0.1 significance level’]

[Using Excel Function: Statistical, TINV(Probability, Deg_freedom) which gives T for which P(|t| > T) = given probability]

p-value = P(t_{n – 1} < tcal) = P(t₂₁ < - 1.2792) = 0.1074

[Using Excel Function: Statistical, TDIST(x, Degrees_freedom, Tails)]

Decision:

Since tcal > tcrit, H₀ is accepted.

[Also confirmed by p-value > α.

Conclusion:

There is not sufficient evidence to support the claim that the new forms take less time to complete than the older forms.

DONE