Question

The university finance department wants to know if the average age of students at their university is greater than the average for other universities. A random sample of student records is taken from the own university (population 1) and a random selection of student ages from other three universities are taken (population 2). A significance level of 0.05 is chosen.

The null and alternative hypotheses are:

?0:

??:

The samples are selected, and the results are:

?1 = 28,7 ?????   ?1 = 5.1 ?????   ?1 = 125

?2 = 24,9 ?????   ?2 = 3.5 ?????   ?2 = 250

	Sample 1	Sample 2
n (size)	125	250
x_bar	28,7	24,9
stdev	5,1	3,5
variance	26,01	12,25
st.err
z
alpha	0,05
zα
p-value

Answer 1

The provided information is:

Sample 1:

Sample 2:

Significance level is 0.05.

\\

The researcher wants to test whether the average age of students at their university is greater than the average age of students from other three universities.

Let represents the population mean age of students at their own university.

Let represents the population mean age of students from other three universities.

The null and the alternative hypothesis formulated as,

The test is right tailed.

Since the larger value of the sample standard deviation is not double in size than the smaller vale of the sample standard deviation, so we compare two means by using pooled method.

The pooled standard deviation is given by the following formula,

The test statistic is given as,

The degrees of freedom is n1 + n2 – 2 = 125+250 -2 = 373.

\\

Using the t-table, the p-value is obtained as,

Since the p-value (0.00001) is less than significance level (0.05), so the null hypothesis is rejected.

Hence, it is concluded that there is a sufficient evidence to support that the average age of students at their university is greater than the average age of students from other three universities.