Question

The average length of a certain species of butterfly is believed to be 0.75 inches. A researcher believes that actually the average length should be higher than what is believed. To test the hypothesis H 0: μ = 0.75 against H a: μ > 0.75, the researcher takes a sample of 36 butterflies and found that the mean from the sample is 0.65 inches and the sample standard deviation s=1.99 inches. What test should you use for this test?a. T-testb. Z-test

Answer 1

Answer:

a. T-test

If we analyze the size for the sample is > 30 but we know the population deviation so is better apply a t test to compare the actual mean to the reference value

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X=0.65$ represent the sample mean  

$s=1.99$ represent the sample standard deviation  

n=36 represent the sample selected  

$\alpha$ significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is higher than 0.75 inches, the system of hypothesis would be:    

Null hypothesis:$\mu \leq 0.75$    

Alternative hypothesis:$\mu > 0.75$    

If we analyze the size for the sample is > 30 but we 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:    

$t=\frac{0.65-0.75}{\frac{1.99}{\sqrt{36}}}=-0.3015$      

We can also calculate the p value given by:

$p_v =P(z>-0.3015)=0.618$    

Conclusion    

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, the true mean is not significantly higher than 0.75.