Question

A marketing study was conducted to compare the mean age of male and femlae purchasers of a certain product. random and independent samples were selected fro both male and female purchasers of the product. it was desired to test to determine i the mean age of all female purchasers exceeds the mean age of all male purchasers.The sample data is shown here:Female: n = 10, sample mean = 50.30, sample standard deviation = 13.215Male: n = 10, sample mean = 39.80, sample standard deviation = 10.0401. Find the rejection region when testing at alpha = 0.05.A. t ≥ 1.734B. t ≥ 2.101C. t ≥ 2.528D. t ≥ 1.330E. None of the above.

Answer 1

Answer:

A. t ≥ 1.734

Step-by-step explanation:

Data given and notation  

$\bar X_{m}=39.80$ represent the mean for the sample male

$\bar X_{f}=50.30$ represent the mean for the sample female

$s_{m}=10.040$ represent the sample standard deviation for the males  

$s_{f}=13.215$ represent the sample standard deviation for the females  

$n_{m}=10$ sample size for the group male  

$n_{f}=10$ sample size for the group female  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for females is higher than the mean for males, the system of hypothesis would be:  

Null hypothesis:$\mu_{f}-\mu_{m}\leq 0$  

Alternative hypothesis:$\mu_{f} - \mu_{m}> 0$  

We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  

$t=\frac{(\bar X_{f}-\bar X_{m})-\Delta}{\sqrt{\frac{s^2_{f}}{n_{f}}+\frac{\sigma^2_{m}}{n_{m}}}}$ (1)

And the degrees of freedom are given by $df=n_m +n_f -2=10+10-2=18$  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

What is the test statistic?

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

$t=\frac{(50.3-39.80)-0}{\sqrt{\frac{13.215^2}{10}+\frac{10.040^2}{16}}}}=2.001$

Critical value

For this case since we have a right tailed test we need to look into the t distribution with 18 degrees of freedom a value that accumulates 0.05 of the area on the right, and on this case:

$t_{crit}=1.734$

And the rejection zone of the null hypothesis would be: A. t ≥ 1.734

For our case our calculated value is higher than the critical value so we have enough evidence to reject the null hypothesis at the 5% of significance.