Question

A training director for a large company has been told that, on completion of the train- ing course, the average score of her trainees on the final evaluation should be 100. Her only concern is whether she will have to be- gin remedial steps to ensure that the popu- lation of trainees is not below standard. She draws a random sample of 10 scores of re- cent trainees: 94, 98, 101, 90, 86, 102, 95, 100, 98, and 92. (a) State the null and alternative hypotheses best suited to the nature of her inquiry.Test the null hypothesis (b) at the .05 level of significance and state your conclu- sions and (c) at the .01 level of significance

Answer 1

Answer:

a) Null hypothesis:$\mu \geq 100$  

Alternative hypothesis:$\mu < 100$  

$t=\frac{95.6-100}{\frac{5.168}{\sqrt{10}}}=-2.692$    

$df=n-1=10-1=9$  

Since is a one side test the p value would be:  

$p_v =P(t_{(9)}$  

b) 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, so we can conclude that the true mean is significantly lower than 100 at 5% of significance

c) 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, so we can conclude that the true mean is NOT significantly lower than 100 at 1% of significance

Step-by-step explanation:

Data given and notation  

We can calculate the sample mean and deviation with thie following formulas:

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

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=95.6$ represent the sample mean

$s=5.168$ represent the sample standard deviation for the sample

$n=10$ sample size  

$\mu_o =100$ 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 true mean is lower than 100, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 100$  

Alternative hypothesis:$\mu < 100$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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:  

$t=\frac{95.6-100}{\frac{5.168}{\sqrt{10}}}=-2.692$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=10-1=9$  

Since is a one side test the p value would be:  

$p_v =P(t_{(9)}$  

Part b: 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, so we can conclude that the true mean is significantly lower than 100 at 5% of significance

Part c

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, so we can conclude that the true mean is NOT significantly lower than 100 at 1% of significance