Answered

Discover new information and insights with the help of IDNLearn.com. Whether it's a simple query or a complex problem, our experts have the answers you need.

Consider the following hypothesis test:
H0: = 18
Ha: ≠ 18
A sample of 48 provided a sample mean = 17 and a sample standard deviation s = 4.5.
If requires, round your answers to two decimal places.
a. Compute the value of the test statistic (to three decimal places.)
b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the p-value.
p-value is between ______and _______
c. What is your conclusion? t = ______

Sagot :

MrRoyalidn

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]

We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.