In the past, the mean age of consumers of a product was 37 years. Recently the product has been drastically changed. In order to determine whether there has been a change in the mean age of the product consumers, a sample of 16 consumers is selected. The mean age in the sample is 31 years and the standard deviation in the sample is 10 years. Let μ = the mean age of the current product consumers.
At a 1% significance level, the null hypothesis is rejected if:_________.

We can calculate the critical value for this case with the degrees of freedom ,we can find a critical value in the t distribution with 15 degrees of freedom who accumulate 0.005 of the area on each tail of the distribution and we got:

[tex] t_{crit}=\pm 2.947[/tex]

And we will reject the null hypothesis is the statistic for this case satisfy this condition:

[tex] |t_{calculated}| > |t_{critical}|= 2.947[/tex]

Step-by-step explanation:

Information given

[tex]\bar X=31[/tex] represent the sample mean

[tex]s=10[/tex] represent the sample standard deviation

[tex]n=16[/tex] sample size

[tex]\mu_o =37[/tex] represent the value to verify

[tex]\alpha=0.01[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value for the test

System of hypotheis

We want to verify if the actual mean age of consumers of a product was 37 years and the system of hypothesis are:

Null hypothesis:[tex]\mu = 37[/tex]

Alternative hypothesis:[tex]\mu \neq 37[/tex]

We can use the following statistic for this case:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

Replacing the info given we got:

[tex]t=\frac{31-37}{\frac{10}{\sqrt{16}}}=-2.4[/tex]

The degree of freedom are:

[tex]df=n-1=16-1=15[/tex]

We can calculate the critical value for this case with the degrees of freedom we can find a critical value in the t distribution with 15 degrees of freedom who accumulate 0.005 of the area on each tail of the distribution and we got:

[tex] t_{crit}=\pm 2.947[/tex]

And we will reject the null hypothesis is the statistic for this case satisfy this condition:

[tex] |t_{calculated}| > |t_{critical}|= 2.947[/tex]