Answer:
The calculated p-value is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & conclude that there is no significant evidence to say that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.
That is, the true mean waiting time is equal to or greater than the 15-minute claim by the taxpayer advocate.
Step-by-step explanation:
For hypothesis testing, we first clearly state our null and alternative hypothesis.
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
In hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the direction of the test.
The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the direction of the test.
For this question, we are to investigate that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.
The null hypothesis would be that there is no significant evidence to say that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate. That is, the true mean waiting time is equal to or greater than the 15-minute claim by the taxpayer advocate.
The alternative hypothesis is that there is significant evidence to suggest that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.
This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim
Mathematically, the null hypothesis is
H₀: μ ≥ 15
The alternative hypothesis is
Hₐ: μ < 15 minutes
To do this test, we will use the z-distribution because the population standard deviation is known.
So, we compute the z-test statistic
z = (x - μ₀)/σₓ
x = sample mean = 13 minutes
μ₀ = the advocate's claim = 15 minutes
σₓ = standard error of the poll proportion = (σ/√n)
where n = Sample size = 50
σ = population standard deviation = 11 minutes.
σₓ = (σ/√n) = (11/√50) = 1.556
z = (13 - 15) ÷ 1.556 = -1.29
checking the tables for the p-value of this z-statistic
p-value (for z = -1.29, at 0.05 significance level, with a one tailed condition) = 0.098525
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 5% = 0.05
p-value = 0.098525
0.098525 > 0.05
Hence,
p-value > significance level
This means that we fail to reject the null hypothesis & conclude that there is no significant evidence to say that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.
That is, the true mean waiting time is equal to or greater than the 15-minute claim by the taxpayer advocate.
Hope this Helps!!!
