Question

Male cardinal birds sing to attract females. An ornithologist wants to test the claim that the true mean duration of songs sung by male cardinals is more than 6 seconds. She takes a SRS of 35 male cardinals and finds a sample mean duration of 6.31 seconds and a sample standard deviation of 1.2 seconds. Which of the following is the appropriate test statistic to test this claim? a. 1.5063 Ob.0.0074 c. 5.465 d. 1.5283 e. 0.0437

Answer 1

Answer:

$t=\frac{6.31-6}{\frac{1.2}{\sqrt{35}}}=1.5283$    

d. 1.5283

Step-by-step explanation:

Data given and notation    

$\bar X=6.31$ represent the sample mean    

$s=1.2$ represent the sample standard deviation    

$n=35$ sample size    

$\mu_o =6$ represent the value that we want to 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 6 seconds, the system of hypothesis are :    

Null hypothesis:$\mu \leq 6$    

Alternative hypothesis:$\mu > 6$    

Since we don't know the population deviation, 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:    

$t=\frac{6.31-6}{\frac{1.2}{\sqrt{35}}}=1.5283$    

d. 1.5283