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
Score 84 94 80 88 77 68 90 74 96 71

Use a 0.01 significance level to test the claim that people do better with the new edition. Assume the standard deviation is 10.4. (a) What kind of test should be used?

A. One-Tailed
B.Two-Tailed
C.It does not matter.

Answer 1

Answer:

We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :  

Null hypothesis:$\mu \leq 75$  

Alternative hypothesis:$\mu > 75$  

A. One-Tailed

$z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189$  

$p_v =P(z>2.189)=0.0143$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Data given and notation  

We have the following data: 84 94 80 88 77 68 90 74 96 71

We can calculate the sample mean with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X=82.2$ represent the sample mean  

$\sigma=10.4$ represent the population standard deviation  

$n=10$ sample size  

$\mu_o =75$ represent the value that we want to test  

$\alpha=0.01$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :  

Null hypothesis:$\mu \leq 75$  

Alternative hypothesis:$\mu > 75$  

A. One-Tailed

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189$  

P-value  

Since is a ONE-TAILED  test the p value would given by:  

$p_v =P(z>2.189)=0.0143$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.