I need help on these n 11 ,12 and 14
1 answer:
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Answer:
Step-by-step explanation:
Hello!
a.
Your study variable is X: "daily income of a parking garage"
The parameter to estimate is the population mean (μ) of the daily income of the parking garage. Since the population mean is a parameter of the normal distribution you need your study variable to be normally distributed to study it. If your study variable hasn't the required distribution, since you have a big enough sample, apply the Central Limit Theorem to approximate the sample mean distribution to normal.
So the assumption you need to make is that X[bar]≈N(μ;δ²/n)
b.
The formula for the Confidence Interval is:
X[bar]±*(S/√n)
I don't have the value of the population standard deviation, so I'll use the approximation with the sample standard deviation.
Sample
n=44
X[bar]= $126
S= $15
X[bar]±*(S/√n)
[126±1.64*(15/√44)]
[122.29;129.71]
c.
With a confidence level of 90%, you can expect the interval [122.29;129.71] will contain the true mean of the daily income of the parking garage.
d.
90% of confidence means that if you were to take 100 samples and calculate a confidence interval for the population mean of the daily income of the parking garages, you'd expect 90 of those intervals to contain the true value of the parameter.
e.
If I were to make a hypothesis test to see if the consultant was right, using the complemental level of significance, since the interval doesn't contain the supposed value of $130, I would reject the null hypothesis and conclude that he wasn't right.
H₀: μ = 130
H₁: μ ≠ 130
α: 0.10
[122.29;129.71]
Interval doesn't include 130 ⇒ to Reject the null hypothesis.
I hope this helps!