The parking garage at the airport has 4750 empty parking spaces and 250 full parking spaces. What percent of the spaces in the g
arage are empty write your answer using a %
1 answer:
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Answer:
a. [36.19;39.21]
b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.
Step-by-step explanation:
Hello!
Your study variable is,
X: "number of users of one server at a time"
The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.
You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:
X[bar]≈N(μ;σ²/n)
a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z
[X[bar] ± * S/√n]
=
= 1.64
[37.7± 1.64* 9.2/√100]
[36.19;39.21]
b. You need to test if the population mean is greater than 35 with a level of significance of 1%.
The hypothesis is:
H₀: μ ≤ 35
H₁: μ > 35
α: 0.01
This is a one-tailed test so you have only one critical level (right tail):
This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.
If the value is less than 2.33 you will support the null hypothesis.
The statistic is:
Z=<u> X[bar] - μ </u>= <u> 37.7 - 35 </u> = 2.93
S/√n 9.2/10
The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.
<u><em>Note: </em></u><em>To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.</em>
I hope it helps!
Answer:
68% of the rates are expected to be betwen 0.718 and 0.762
95% of the rates are expected to be betwen 0.696 and 0.784
99.7% of the rates are expected to be betwen 0.674 and 0.806
Step-by-step explanation:
Check for conditions
For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:
a) Independence : we assume that the random sample of 400 students each student is independent from the other.
b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.
c) np= 400*0.74= 296>10
n(1-p) = 400*(1-0.74)=104>10
So then we have all the conditions satisfied.
Solution to the problem
For this case we know that the distribution for the population proportion is given by:
So then:
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
68% of the rates are expected to be betwen 0.718 and 0.762
95% of the rates are expected to be betwen 0.696 and 0.784
99.7% of the rates are expected to be betwen 0.674 and 0.806
