Question

In order to monitor the ecological health of a particular​ region, various measurements are recorded at different times. The bottom temperatures are recorded and the mean of 30.385​°C is obtained for 61 temperatures recorded on 61 different days. Assuming that sigmaequals1.7​°C, test the claim that the population mean is greater than 30.0​°C. Use a 0.05 significance level and the displayed results.

Answer 1

Answer:

$z=\frac{30.385-30}{\frac{1.7}{\sqrt{61}}}=1.769$    

$p_v =P(z>1.769)=0.0384$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that mean is significant higher than 30 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=30.385$ represent the sample mean

$\sigma=1.7$ represent the population standard deviation for the sample  

$n=61$ sample size  

$\mu_o =30$ 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 is higher than 30C, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 30$  

Alternative hypothesis:$\mu > 30$  

If we analyze the size for the sample is > 30 and we know the population deviation so 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:  

$z=\frac{30.385-30}{\frac{1.7}{\sqrt{61}}}=1.769$    

P-value

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

$p_v =P(z>1.769)=0.0384$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that mean is significant higher than 30 at 5% of signficance.