Respuesta :
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] ± [tex]Z_{1-\alpha /2}[/tex] * S/√n]
[tex]Z_{1-\alpha /2}[/tex] = [tex]Z_{0.95}[/tex] = 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):
[tex]Z_{1\alpha } = Z_{0.99} = 2.33[/tex]
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= X[bar] - μ = 37.7 - 35 = 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.
Note: To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.
I hope it helps!
The 90% confidence interval is [36.19, 39.21]. The population mean of users that are connected at the same time is greater than 35.
What are null hypotheses and alternative hypotheses?
In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.
The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.
In order to ensure efficient usage of a server, it is necessary to estimate the mean number of concurrent users.
According to records, the average number of concurrent users at 100 randomly selected times is 37.7, with a standard deviation of 9.2.
a) With this approximation, you can construct the 90% confidence interval the approximate z will be
[tex]\mu \pm z_{0.90} * \dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]z_{0.90} = 1.64[/tex]
Then we have
[tex]37.7 \pm 1.64*\dfrac{9.2}{\sqrt{100}}\\\\[/tex]
[36.19, 39.21]
b) You need to test if the population mean is greater than 35 with a level of significance of 1%.
[tex]\rm H_o : \mu\leq 35\\\\H_a : \mu > 35\\\\ \alpha =0.01[/tex]
This is a one-tailed test so you have only one critical level
[tex]z_{1\alpha } = z_{0.99} = 2.33[/tex]
This means that if the value of the statistic is equal to or greater than 2.33 you will reject the null hypothesis.
The statistic will be
[tex]z = \dfrac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\\\\\\z = \dfrac{37.7-35}{\frac{9.2}{\sqrt{10}}} = 2.93[/tex]
The value of z > 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.
More about the null hypotheses and alternative hypotheses link is given below.
https://brainly.com/question/9504281
