Guys i’m begging someone help me please!!!!
2 answers:
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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!