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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 :

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]

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