Answer:
Step-by-step explanation:
Given that:
For PCC
The sample size [tex]n_1[/tex] = 25
sample mean [tex]\overline x_1[/tex] = 75
standard deviation [tex]s_1[/tex] = 17.50
For ELAC
The sample size [tex]n_2[/tex] = 20
Sample mean [tex]\overline x_2 = 89[/tex]
Standard deviation [tex]s_2[/tex] = 14.40
Significance level = 0.05
The null hypothesis:
[tex]H_o : \mu_1 =\mu_2[/tex]
The alternative hypothesis;
[tex]H_1 : \mu_1< \mu_2[/tex]
Since the population standard deviation are synonymous pooled standard deviation; Then:
[tex]sp = \sqrt {\dfrac {(n_1-1)s_1^2)+(n_2-1)s_2^2 }{n_1+n_2-2}[/tex]
[tex]sp = \sqrt {\dfrac {(25-1)17.50^2)+(20-1)14.40^2 }{25+20-2}[/tex]
[tex]sp = 16.20[/tex]
The test statistics can be computed as:
[tex]t = \dfrac{\overline x_1 -\overline x_2}{sp \times \sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2} }}[/tex]
[tex]t = \dfrac{75-89}{16.20 \times \sqrt{\dfrac{1}{25} + \dfrac{1}{20} }}[/tex]
[tex]t = -2.88[/tex]
The p-value [tex]= P(t_{n_1+n_2-1} <t)[/tex]
[tex]= P(t_{43} <-2.88)[/tex]
= 0.0031
Decision Rule: To reject the null hypothesis if the p-value is less than the significance level
Conclusion: There is sufficient evidence to conclude that the mean amount spent by all PCC students is less than the mean amount spent by all ELAC students.
