Question

2. In an industrial training program, students have been averaging about 64 points on a standardized test. The lecture system was replaced by teaching machines with a lab instructor. There was some doubt as to whether the scores would decrease, increase, or stay the same. A sample of n = 60 students using the teaching machines was tested, resulting in a mean of 68 and a standard deviation of 12. Perform a hypothesis test to see if scores would decrease, increase, or stay the same. Use α = 0.05. Be sure to:
1. State your hypotheses.
2. Find the value of the Test Statistic.
3. Find the p-value
4. State your decision (Reject or not)
5. State your conclusion.

Answer 1

Answer:

Case I

Null hypothesis:$\mu = 64$  

Alternative hypothesis:$\mu \neq 64$  

$t=\frac{68-64}{\frac{12}{\sqrt{60}}}=2.582$  

$df=n-1=60-1=59$  

Since is a two sided  test the p value would given by:  

$p_v =2*P(t_{(59)}>2.582)=0.012$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.  

We can say that at 5% of significance the true mean is different from 64.

Case II

Null hypothesis:$\mu \leq 64$  

Alternative hypothesis:$\mu > 64$

The statistic not changes but the p value does and we have:

$p_v =P(t_{(59)}>2.582)=0.006$  

And we reject the null hypothesis on this case.

So we can conclude that the true mean is significantly higher than 64 at 5% of singnificance

Step-by-step explanation:

Data given and notation  

$\bar X=68$ represent the sample mean  

$s=12$ represent the sample standard deviation  

$n=60$ sample size  

$\mu_o =64$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t 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 population mean is different from 64 the system of hypothesis are :  

Null hypothesis:$\mu = 64$  

Alternative hypothesis:$\mu \neq 64$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-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:  

$t=\frac{68-64}{\frac{12}{\sqrt{60}}}=2.582$  

P-value  

We need to calculate the degrees of freedom first given by:  

$df=n-1=60-1=59$  

Since is a two sided  test the p value would given by:  

$p_v =2*P(t_{(59)}>2.582)=0.012$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.  

We can say that at 5% of significance the true mean is different from 64.

Now let's assume that we want to see if the mean is significantly higher than 64

Null hypothesis:$\mu \leq 64$  

Alternative hypothesis:$\mu > 64$

The statistic not changes but the p value does and we have:

$p_v =P(t_{(59)}>2.582)=0.006$  

And we reject the null hypothesis on this case.

So we can conclude that the true mean is significantly higher than 64 at 5% of singnificance