With is greater 5.6 or 5.06 and why
2 answers:
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Answer:
Step-by-step explanation:
Hello!
The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.
For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.
The established variable has a normal distribution with a standard deviation of σ= 2 min.
From the taken sample an average time difference of X[bar]= 0.77 was obtained.
If the app is correct, the true mean should be around cero, symbolically: μ=0
a. The hypotheses are:
H₀:μ=0
H₁:μ≠0
b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.
c.
d. This hypothesis test is two-tailed and so is the p-value.
p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058
e. 90% CI
X[bar] ±
0.77 ± 1.645 *
[0.098;1.442]
I hope this helps!