Question

question 2 (10 total marks; 2 marks each)
the mean score on the final exam for a statistics course is advertised to be 75. there is reason to believe that the average score is higher. ten randomly selected students who took the final exam and their scores are shown in the table below. use a 1% significance level to test the claim that students did better than 75.
student a b c d e f g h i j
test score 85 83 94 96 73 92 70 68 76 87
a. state the hypotheses.
solution:
b. find the test statistic (round to 4 decimals)
solution:

Explanation:

Step1: State the hypotheses

The null hypothesis $H_0:\mu = 75$ and the alternative hypothesis $H_1:\mu>75$.

Step2: Calculate the sample mean $\bar{x}$

$\bar{x}=\frac{85 + 83+94+96+73+92+70+68+76+87}{10}=\frac{834}{10}=83.4$

Step3: Calculate the sample standard - deviation $s$

First, calculate the sum of squared differences from the mean:
\[

$$\begin{align*} \sum_{i = 1}^{n}(x_i-\bar{x})^2&=(85 - 83.4)^2+(83 - 83.4)^2+(94 - 83.4)^2+(96 - 83.4)^2+(73 - 83.4)^2+(92 - 83.4)^2+(70 - 83.4)^2+(68 - 83.4)^2+(76 - 83.4)^2+(87 - 83.4)^2\\ &=(1.6)^2+(- 0.4)^2+(10.6)^2+(12.6)^2+(-10.4)^2+(8.6)^2+(-13.4)^2+(-15.4)^2+(-7.4)^2+(3.6)^2\\ &=2.56 + 0.16+112.36+158.76+108.16+73.96+179.56+237.16+54.76+12.96\\ &=930.4 \end{align*}$$

\]
The sample standard - deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{930.4}{9}}\approx10.178$

Step4: Calculate the test statistic $t$

The test statistic for a one - sample $t$-test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\mu = 75$, $\bar{x}=83.4$, $s\approx10.178$, and $n = 10$.
$t=\frac{83.4 - 75}{10.178/\sqrt{10}}=\frac{8.4}{10.178/3.1623}\approx\frac{8.4}{3.2186}\approx2.6100$

Answer:

a. $H_0:\mu = 75$, $H_1:\mu>75$
b. $t\approx2.6100$