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 the hypotheses

The null hypothesis $H_0:\mu = 71$ and the alternative hypothesis $H_1:\mu
eq71$.

Step2: Calculate the test - statistic

The formula for the t - test statistic in a one - sample t - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$. Given $\bar{x} = 73$, $\mu = 71$, $s = 12$, and $n = 23$.
\[t=\frac{73 - 71}{12/\sqrt{23}}=\frac{2}{12/\sqrt{23}}=\frac{2\sqrt{23}}{12}\approx 0.7986\]

Step3: Find the critical values

The degrees of freedom is $df=n - 1=23-1 = 22$. For a two - tailed test with $\alpha=0.01$, the critical values are $t_{\alpha/2,df}=t_{0.005,22}$. From the t - distribution table, $t_{0.005,22}\approx\pm 2.8188$. Since $- 2.8188<0.7986<2.8188$, we fail to reject the null hypothesis.

Step4: Calculate the p - value

For a two - tailed t - test with $t = 0.7986$ and $df = 22$, using a t - distribution calculator or software, the p - value is $P(|t|>0.7986)=2P(t>0.7986)$. Looking up in the t - distribution table or using software, $P(t>0.7986)\approx0.216$, so the p - value is $2\times0.216 = 0.432$. Since $p - value=0.432>0.01$, we fail to reject the null hypothesis.

Answer:

a. $H_0:\mu = 71$, $H_1:\mu
eq71$
b. $t\approx0.7986$
c. Fail to reject $H_0$ since $-2.8188<0.7986<2.8188$
d. Fail to reject $H_0$ since $p - value = 0.432>0.01$