Answer:
(a) [tex]t= -1.540[/tex]
(b) [tex]0.10 < p < 0.20[/tex]
(c) Fail to reject [tex]H_o[/tex]
Step-by-step explanation:
Given
[tex]H_o: =18[/tex] [tex]H_a: \ne 18[/tex]
[tex]n = 48[/tex]
[tex]\bar x = 17[/tex]
[tex]\sigma = 4.5[/tex]
Solving (a): The test statistic
This is calculated as:
[tex]t= \frac{\bar x - \mu_o}{\sigma/\sqrt n}[/tex]
So, we have:
[tex]t= \frac{17 - 18}{4.5/\sqrt{48}}[/tex]
[tex]t= \frac{- 1}{4.5/6.93}[/tex]
[tex]t= \frac{- 1}{0.6493}[/tex]
[tex]t= -1.540[/tex] --- approximated
Solving (b): Range of p value
First, calculate the degree of freedom (df)
[tex]df = n - 1[/tex]
[tex]df = 48 - 1[/tex]
[tex]df = 47[/tex]
Using:
[tex]\alpha = 0.05[/tex] --- significance level
The p value at: [tex]df = 47[/tex] is:
[tex]p = 0.065133[/tex]
and the range is:
[tex]0.05 * 2 < p < 2 * 0.10[/tex]
[tex]0.10 < p < 0.20[/tex]
Solving (c): The conclusion
Compare the p value to the level of significance value
We have:
[tex]p = 0.065133[/tex]
[tex]\alpha = 0.05[/tex]
By comparison:
[tex]p > \alpha[/tex]
because:
[tex]0.065133 > 0.05[/tex]
Hence, the conclusion is: fail to reject [tex]H_o[/tex]
