Response:
a) Critical value = -1.4052
Since we are verifying whether the mean time is under 10 minutes, the critical region is defined by
z < -1.4052
b) Option C is accurate.
Dispute the assertion that the average time is 10 minutes due to the test statistic being lower than the critical threshold.
Thus, it indicates the mean time is significantly under 10 minutes.
Step-by-step breakdown:
a) Using z-distribution, the critical value is derived based on the confidence level of the conducted test. Since the hypothesis test is one-sided (determining if the average time is significantly less than 10 minutes)
P(z < Critical value) = 0.08
Based on z-tables, critical value is -1.4052
Given the direction of interest regarding mean time being less than 10 minutes, the critical region is
z < -1.4052
b) Initially, we establish the null and alternative hypotheses
The null hypothesis implies insufficient evidence to claim that the mean time is below 10 minutes.
Conversely, the alternative hypothesis posits that there is substantial evidence indicating the mean time is under 10 minutes.
To perform this hypothesis assessment now, we need to calculate the test statistic
Test statistic = (x - μ)/σₓ
x = mean sample = (Σx/N)
The observed data is
1.2, 2.8, 1.5, 19.3, 2.4, 0.7, 2.2, 0.7, 18.8, 6.1, 6, 1.7, 29.1, 2.6, 0.2, 10.2, 5.1, 0.9, 8.2
Σx = 119.7
N = Size of sample = 19
x = sample mean = (119.7/19) = 6.3
μ = standard for comparison = 10 minutes
σₓ = standard error = (σ/√N)
where N = Size of sample = 19
σ = √[Σ(x - xbar)²/N]
x = each data point
xbar = mean = 6.3
N = Sample size = 19
Σ(x - xbar)² = 1122.74
σ = (√1122.74/19) = 7.687
σₓ = (7.687/√19) = 1.7635
Test statistic = (x - μ)/σₓ
Test statistic = (6.3 - 10)/1.7635
= -2.098 = -2.10
z = -2.10 falls within the rejection area (z < -1.4052), hence, we reject the null hypothesis in favor of the claim, asserting the mean time is significantly below 10 minutes.
