Question

According to the norms established for a reading comprehension test, eighth-graders should average 83.2 with a standard deviation of 8.6. If 45 randomly selected eighth graders from a certain school district averaged 86.7 with a standard deviation of 8.6, test the null hypothesis LaTeX: \muμ= 83.2 against the alternative hypothesis LaTeX: \muμ> 83.2 at the .01 level of significance. And then decide whether there is evidence to support the superintendent's claim that her eighth graders are above average.

Answer 1

Answer:

$t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73$    

$p_v =P(t_{44}>2.73)=0.0045$    

If we compare the p value and a significance level given $\alpha=0.01$ 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 population mean is signficantly higher than 83.2 at 1% of significance.

Step-by-step explanation:

Data given and notation    

$\bar X=86.7$ represent the average life for the sample    

$s=8.6$ represent the population standard deviation    

$n=45$ sample size    

$\mu_o =83.2$ represent the value that we want to test    

$\alpha=0.01$ 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 apply a two tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  $\mu \leq 83.2$  

Alternative hypothesis :$\mu > 83.2$  

Compute the test statistic  

The statistic for this case is given by:  

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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:    

$t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73$    

Give the appropriate conclusion for the test  

First we need to calculate the degrees of freedom given by:

$df=n-1=45-1=44$

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

$p_v =P(t_{44}>2.73)=0.0045$    

Conclusion    

If we compare the p value and a significance level given $\alpha=0.01$ 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 population mean is signficantly higher than 83.2 at 1% of significance.