Step-by-step explanation:
Mean of the test of programming ability was 160 in the past.
Again now, twenty-five job applicants are randomly selected from one university and they produce a mean score and standard deviation of 183 and 12.
So in this case, the null hypothesis will be,
[tex]H_o: \mu = 160[/tex]
and alternate hypothesis will be,
[tex]H_a: \mu > 160[/tex]
Mean of the sample = [tex]\overline{x}[/tex] = 183
Standard deviation of the sample = [tex]\sigma[/tex] = 12
sample size = n = 25
Using t distribution,
[tex]t=\dfrac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]=\dfrac{183-160}{\frac{12}{\sqrt{25}}}[/tex]
[tex]=9.58[/tex]
P-value =[tex]P(t > 9.58)\approx 0[/tex]
As the obtained value is less than 0.05 or 5% level of significance, so we have to reject the null hypothesis.
