Respuesta :
Answer:
[tex]H_0:\mu = 45[/tex] and [tex]H_a:\mu < 45[/tex]
[tex]T = 34.88[/tex]
The null hypothesis is not rejected
Step-by-step explanation:
Given
Random sample of 50 commuter times
Solving (a): The null and alternate hypothesis
From the question, a previous census shows that:
[tex]\mu =45[/tex] and [tex]\sigma= 9.1[/tex]
This is the null hypothesis i.e.
[tex]H_0:\mu = 45[/tex]
Reading further, a new test was to show whether the average has decreased.
This is the alternate hypothesis, and it is represented as:
[tex]H_a:\mu < 45[/tex]
Solving (b) and (c): The test statistic (T).
This is calculated using:
[tex]T = \frac{\mu}{\sigma/\sqrt n}[/tex]
In this case:
[tex]n = 50[/tex] [tex]\mu =45[/tex] and [tex]\sigma= 9.1[/tex]
So:
[tex]T = \frac{45}{9.1/\sqrt{50}}[/tex]
[tex]T = \frac{45}{9.1/7.07}[/tex]
[tex]T = \frac{45}{1.29}[/tex]
[tex]T = 34.88[/tex]
Solving (c): Accept or reject null hypothesis
[tex]\alpha = 0.05[/tex]
First calculate the degrees of freedom (df)
[tex]df = n - 1 = 50 - 1[/tex]
[tex]d f = 49[/tex]
From the t distribution table,
At: [tex]d f = 49[/tex], [tex]\alpha = 0.05[/tex] and [tex]T = 34.88[/tex]
the t score is:
[tex]p = 1.676[/tex]
[tex]p > 0.05[/tex]
So, the null hypothesis is not rejected