Question

A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. The mean score on the final exam for a course using the old edition is 75. Ten randomly selected people who used the new text take the final exam. Their scores are shown in the table below. Person A B C D E F G H I J Test Score 92 83 93 74 85 97 88 70 69 79 Use a 0.010.01 significance level to test the claim that people do better with the new edition. Assume the standard deviation is 10.5. (Note: You may wish to use statistical software.)

Answer 1

Answer:

The null and alternative hypotheses are:

$H_{0}:\mu=75$

$H_{a} : \mu>75$

Under the null hypothesis, the test statistic is:

$t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}} }$

Where:

$\bar{x} =83$ is the sample mean

$s=9.8206$ is the sample standard deviation

$n=10$ is the sample size

$\therefore t= \frac{83-75}{\frac{9.8206}{\sqrt{10}} }$

         $=2.58$

Now, we can find the right tailed t critical value at 0.01 significance level for df = n-1 = 10 - 1 = 9 using the t distribution table. The t critical value is given below:

$t_{critical} =2.821$

Since the test statistic is less than the t critical value, we therefore, fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that the people do better with the new edition.