Question

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Answer 1

Answer:

$p_v =P(t_{(9)}>1.978)=0.0397$  

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 say that the true mean is higher than 65 at 5 % of significance.

Step-by-step explanation:

Data given and notation  

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

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

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

$\bar X=67$ represent the sample mean  

$s=3.197$ represent the sample standard deviation for the sample  

$n=10$ sample size  

$\mu_o =65$ 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 mean score is higher than 65, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 65$  

Alternative hypothesis:$\mu > 65$  

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:  

$t=\frac{67-65{\frac{3.197}{\sqrt{10}}}=1.978$    

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 right tailed test the p value would be:  

$p_v =P(t_{(9)}>1.978)=0.0397$  

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 say that the true mean is higher than 65 at 5 % of significance.

Other interpretation is:  If the null hypothesis is true, then there is a 0.0397 probability (3.96%) that the sample mean is 65 or more.