Question

We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys walk is approximately normally distributed with a standard deviation of 2.5 months. A random sample of 25 boys has a mean of 19.2 months. Which of the following statements are correct?
I. This finding is significant for a two-tailed test at .05.
II. This finding is significant for a two-tailed test at .01.
III. This finding is significant for a one-tailed test at .01.

a. I only
b. II only
c. III only
d. II and III only
e. I and III only

Answer 1

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

$\bar X=19.2$ represent the battery life sample mean    

$\sigma=2.5$ represent the population standard deviation    

$n=25$ sample size    

$\mu_o =18$ represent the value that we want to test    

$\alpha$ 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)    

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:$\mu = 18$    

Alternative hypothesis:$\mu \neq 18$    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:$\mu \leq 18$    

Alternative hypothesis:$\mu > 18$  

Since we know the population deviation, 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".    

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

$z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4$    

4) P-value    

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

$df=n-1=25-1=24$  

Since is a two tailed test for parts I and II, the p value would be:    

$p_v =2*P(t_{(24)}>2.4)=0.0245$

And for part III since we have a one right tailed test the p value is:

$p_v =P(t_{(24)}>2.4)=0.0122$

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the $p_v$. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the $p_v >\alpha$. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the $p_v >\alpha$. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only