Answer:
Step-by-step explanation:
In the past, mean of age of employees
i.e. [tex]\mu = 40[/tex]
Recently sample was taken
n = sample size = 60
Mean of sample = 45
Std dev of sample s = 16
[tex]H_0: \bar x = 40\\H_a: \bar x >40[/tex]
(Right tailed test)
Since only population std deviation is known we can use t test only
Std error = [tex]\frac{s}{\sqrt{n} } \\=\frac{16}{\sqrt{61} } \\=2.049[/tex]
Mean difference = 45-40 =5
Test statistic t=[tex]\frac{5}{2.049} \\=2.441[/tex]
df = 60
p value =0.008739
Since p < 0.05 we reject null hypothesis
The mean age has increased.