Question

(2) In an experiment designed to study effects of illumination level on task performance2, subjects were required to insert a fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each observation is the time each subject took for task completion. (a) Estimate the true mean difference in completion time with 95% confidence. Interpret (b) Does the data indicate that the higher level of illumination decreases the time for task completion? Conduct hypothesis test (c) State the kind of error could have been made in context of the problem. tasks di 1 25.85 18.23 -7.62 2 28.84 20.84 -8.00 3 32.05 22.96 -9.09 4 25.74 19.68 6.06 5 20.89 19.50 -1.39 6 41.05 24.98 -16.07 7 25.01 16.61-8.40 8 24.96 16.07 -8.89 9 27.47 24.59 2.88 subject black white 2 4 5 6 7 rbind (dbar,sd.d,se.d) dbar7.600000 sd.d 4.177912 se.d 1.392637

Answer 1

Answer a)

Step 1: Find ?/2
Level of Confidence = 95%
? = 100% - (Level of Confidence) = 5%
?/2 = 2.5% = 0.025

Step 2: Find t_?/2
Calculate t_?/2 by using t-distribution with degrees of freedom (DF) as n - 1 = 9 - 1 = 8 and ?/2 = 0.025 as right-tailed area and left-tailed area.

t_?/2 = 2.306 (Obtained using t distribution table. Screenshot attached)

Step 3: Calculate Confidence Interval
Confidence Formula: [d? - t_?/2•(s_d/?n) , d? + t_?/2•(s_d/?n)]
t_?/2 = 2.306, s_d = 4.177912, n = 9, d? = -7.60

Lower Bound = d? - t_?/2•(s_d/?n) = -7.6 - (2.306)(4.177912/?9) = -10.811
Upper Bound = d? + t_?/2•(s_d/?n) = -7.6 + (2.306)(4.177912/?9) = -4.389
Confidence Interval = (-10.811, -4.389)

Interpretation:

Since we do not know if the confidence interval (-10.811, -4.3886479412) contains the true mean difference ?_d or not, we are only 95% confident that (-10.811, -4.389) contains the true mean difference.

Answer b)

Step 1: Formulate hypothesis

H₀: ?_d = 0 (There is no difference in completion time)
H₁: ?_d < 0 (The completion time decreased, with higher level of illumination)

Step 2: Input ? (level of significance of hypothesis test).

? = 0.05 (Note: ? = level of significance of hypothesis test = probability of making Type I error.)

Step 3: Calculate Test Statistic.

d? (sample mean) = -7.60
s_d = 4.177912
n (sample size) = 9

          (Note: From Step 1, we have H₀: ?_d = 0; therefore set ?_d = 0)

d- test statistic t -

Test statistic t = (-7.60-0)/(4.177912/SQRT(9))

Test statistic t = -5.4573

Step 4: Find Critical Value and Rejection Region

t_? is the t-score corresponding to the left-tailed area. Degree of freedom = 9 - 1 = 8

Critical Value is -1.860 (Obtained using t distribution table. Screenshot attached)

Rejection Region: Reject Null Hypothesis If Test statistic < Critical Value

Step 5: Make Decision

Test statistic t = -5.4573

Critical Value is -1.860

In this case, test statistic (-5.4573) is less Critical Value (-1.860) so we reject null hypothesis.

Conclusion: There is sufficient evidence to support the claim that the higher level of illumination decreases the time for task completion.

Answer c)

In statistical hypothesis testing, a type I error is the rejection of a true null hypothesis while a type II error is failing to reject a false null hypothesis. In this case, as we have rejected the null hypothesis, so Type I error could have been made in context of problem.