Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.