Respuesta :
Answer:
At a 0.05 level of significance, it can be concluded that the mean age is significantly different from 24.
Step-by-step explanation:
For hypothesis testing, we first clearly state our null and alternative hypothesis.
The null hypothesis is that the mean age of students isn't significantly different from 24.
And the alternative hypothesis is that the mean age of students is significantly different from 24.
Mathematically, the null hypothesis is
H₀: μ₀ = 24
The alternative hypothesis is
Hₐ: μ₀ ≠ 0.49
To do this test, we will use the z-distribution because we have information about the population standard deviation.
The population of ages is normal with a standard deviation of 2 years.
So, we compute the z-test statistic
z = (x - μ)/σₓ
x = 25
μ = μ₀ = 24
σₓ = standard error of the mean = (σ/√n)
σ = 2
where n = Sample size = 16
σₓ = (2/√16) = 0.50
z = (25 - 24) ÷ 0.50
z = 2.00
checking the tables for the p-value of this z-statistic
p-value (for z = 2.00, at 0.05 significance level, with a two tailed condition) = 0.0455
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 5% = 0.05
p-value = 0.0455
0.0455 < 0.05
Hence,
p-value < significance level
This means that we reject the null hypothesis & accept the alternative hypothesis and we conclude that there is significant evidence to show that the mean age of students is significantly different from 24.
Hope this Helps!!!
Using the z-distribution, it is found that since the p-value of the test is of 0.0228 < 0.05, we have that:
At a .05 level of significance, it can be concluded that the mean age is significantly different from 24.
At the null hypothesis, it is tested if the average age of students is not significantly different of 24 years, that is:
[tex]H_0: \mu = 24[/tex]
At the alternative hypothesis, it is tested if it is different of 24 years, hence:
[tex]H_1: \mu \neq 24[/tex]
We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- [tex]\sigma[/tex] is the standard deviation of the sample.
- n is the sample size.
In this problem, the values of the parameters are: [tex]\overline{x} = 25, \mu = 24, \sigma = 2, n = 16[/tex].
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{25 - 24}{\frac{2}{\sqrt{16}}}[/tex]
[tex]z = 2[/tex]
The p-value of the test is the probability of finding a sample mean above 25 years, hence it is 1 subtracted by the p-value of z = 2.
- Looking at the z-table, z = 2 has a p-value of 0.9772.
1 - 0.9772 = 0.0228.
Since the p-value of the test is of 0.0228 < 0.05, we have that:
At a .05 level of significance, it can be concluded that the mean age is significantly different from 24.
A similar problem is given at https://brainly.com/question/14639462
