Question

Suppose you read online that children first count to 10 successfully when they are 32 months old, on average. You perform a hypothesis test evaluating whether the average age at which gifted children first count to 10 is different than the general average of 32 months. What is the p-value of the hypothesis test? Choose the closest answer.

Answer 1

Answer:

$p_v =2*P(t_{(35)}$  

Step-by-step explanation:

Assuming this info from R

hist(gifted$count)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  

##   21.00   28.00   31.00   30.69   34.25   39.00

## Sd

## [1] 4.314887

Data given and notation  

$\bar X=30.69$ represent the mean  

$s=4.3149$ represent the sample standard deviation

$n=36$ sample size  

$\mu_o =32$ 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different than 32, the system of hypothesis would be:  

Null hypothesis:$\mu = 32$  

Alternative hypothesis:$\mu \neq 32$  

If we analyze the size for the sample is > 30 but we don't 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:  

$t=\frac{30.69-32}{\frac{4.3149}{\sqrt{36}}}=-1.822$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=36-1=35$  

Since is a two sided test the p value would be:  

$p_v =2*P(t_{(35)}$