Respuesta :
Answer:
[tex]t=\frac{54-58}{\frac{6}{\sqrt{70}}}=-5.58[/tex]
Now we can calculate the degrees of freedom:
[tex] df =n-1= 70-1=69[/tex]
And the p value would be given by this probability taking in count the bilateral test:
[tex]p_v =2*P(t_{69}<-5.58)=4.38x10^{-7}[/tex]
Since the p value is lower than the significance level provided we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 58
Step-by-step explanation:
Information given
[tex]\bar X=54[/tex] represent the mean age for the CEOs
[tex]s=6[/tex] represent the sample deviation
[tex]n=70[/tex] sample size
[tex]\mu_o =58[/tex] represent the value to verify
[tex]\alpha=0.01[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if mean age of all CEOs of medium-sized companies in the United States is different from 58 years, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 58[/tex]
Alternative hypothesis:[tex]\mu \neq 58[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{54-58}{\frac{6}{\sqrt{70}}}=-5.58[/tex]
Now we can calculate the degrees of freedom:
[tex] df =n-1= 70-1=69[/tex]
And the p value would be given by this probability taking in count the bilateral test:
[tex]p_v =2*P(t_{69}<-5.58)=4.38x10^{-7}[/tex]
Since the p value is lower than the significance level provided we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 58
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 58
For the alternative hypothesis,
µ ≠ 58
This is a two tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 70
Degrees of freedom, df = n - 1 = 70 - 1 = 69
t = (x - µ)/(s/√n)
Where
x = sample mean = 54
µ = population mean = 58
s = samples standard deviation = 6
t = (54 - 58)/(6/√70) = - 5.58
We would determine the p value using the t test calculator. It becomes
p < 0.0000
Since alpha, 0.01 > than the p value, then we would reject the null hypothesis. Therefore, at a 1% significance level, we can conclude that the current mean age of all CEOs of medium-sized companies in the United States is different from 58 years.
