Question

question 1 (10 total marks; 2 marks each)
statistics final exams have an average score of 71. test the claim using an alternative hypothesis that the mean score is different from 71. sample statistics includes n = 23, $\bar{x}$=73, and s = 12. use a 1% significance level.

a. state the hypotheses.
solution:

b. find the test statistic (round to 4 decimals)
solution:

c. use the critical - value method to test the hypothesis.
solution:

d. use the p - value method to test the hypothesis.
solution:

Explanation:

Step1: State null and alternative hypotheses

$H_0:\mu = 71$, $H_1:\mu
eq71$

Step2: Calculate test - statistic

The formula for the t - test statistic in a one - sample t - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$.
Substitute $\bar{x} = 73$, $\mu = 71$, $s = 12$, and $n = 23$ into the formula:
$t=\frac{73 - 71}{12/\sqrt{23}}=\frac{2}{12/\sqrt{23}}\approx0.7906$

Step3: Determine degrees of freedom and critical values

The degrees of freedom $df=n - 1=23-1 = 22$.
For a two - tailed test with $\alpha=0.01$, the critical values are $t_{\alpha/2,df}=t_{0.005,22}=\pm 2.8188$.
Since $- 2.8188<0.7906<2.8188$, we fail to reject the null hypothesis.

Step4: Calculate p - value

For a two - tailed t - test with $t = 0.7906$ and $df = 22$, using a t - distribution table or software, the p - value is $P(|t|>0.7906)=2P(t>0.7906)$.
$P(t>0.7906)\approx0.219$, so the p - value is approximately $2\times0.219 = 0.438$.
Since $p=0.438>0.01$, we fail to reject the null hypothesis.

Answer:

a. $H_0:\mu = 71$, $H_1:\mu
eq71$
b. $t\approx0.7906$
c. Fail to reject $H_0$ since $-2.8188 < 0.7906<2.8188$
d. Fail to reject $H_0$ since $p = 0.438>0.01$