Respuesta :
Answer:
Option b) significantly different from 24
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 24
Sample mean, [tex]\bar{x}[/tex] = 25
Sample size, n = 25
Alpha, α = 0.05
Population standard deviation, σ = 2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]
Now, we calculate the p-value from the standard normal table.
P-value = 0.0455
Since the p-value is less than the significance level, we fail to accept the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that mean age is significantly different from 24.
At 5% level of significance, it can be concluded that the mean age is B. Significantly different from 24.
How to explain the confidence level?
From the information given, the university has had an average age of 25 years. The critical value at 5% level with the degree of freedom is 1.753 while the test statistic will be:
= (25 - 24)/2 × ✓26
= 1/2 × 4 = 2
Since the test statistic is greater than 1.753, one should reject the null hypothesis. Therefore, it shows that that the mean age is significantly different from 24.
Learn more about significance level on:
https://brainly.com/question/4599596
