Question

In 1960, census results indicated that the age at which men in a certain region first married had a mean of 24.5 years. It is widely suspected that young people today are waiting longer to get married. We want to find out if the mean age of first marriage has increased since then.
We plan to test our hypothesis by selecting a random sample of 40 men who married for the first time last year.
The men in our sample married at an average age of 25.3 years, with a standard deviation of 5.4 years. That results in a t-statistic of 0.937 What is the P-value for this?

Answer 1

Answer:

$t=\frac{25.3-24.5}{\frac{5.4}{\sqrt{40}}}=0.937$    

The degrees of freedom are given by:

$df=n-1=40-1=39$  

And the p value would be given by:

$p_v =P(t_{(39)}>0.937)=0.177$  

Step-by-step explanation:

Information given

$\bar X=25.3$ represent the sample mean

$s=5.4$ represent the sample standard deviation

$n=40$ sample size  

$\mu_o =24.5$ represent the value to verify

t would represent the statistic  

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is higher than 24.5, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 24.5$  

Alternative hypothesis:$\mu > 24.5$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the info given we got:

$t=\frac{25.3-24.5}{\frac{5.4}{\sqrt{40}}}=0.937$    

The degrees of freedom are given by:

$df=n-1=40-1=39$  

And the p value would be given by:

$p_v =P(t_{(39)}>0.937)=0.177$