Answer:
No these these result do not differ at 95% confidence level
Step-by-step explanation:
From the question we are told that
The first concentrations is [tex]c _1= 30.0 \ g/m^3[/tex]
The second concentrations is [tex]c _2 = 52.9 \ g/m^3[/tex]
The first sample size is [tex]n_1 = 32[/tex]
The second sample size is [tex]n_2 = 32[/tex]
The first standard deviation is [tex]\sigma_1 = 30.0 [/tex]
The first standard deviation is [tex]\sigma_1 = 29.0 [/tex]
The mean for Turnpike is [tex]\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125[/tex]
The mean for Tunnel is [tex]\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531[/tex]
The null hypothesis is [tex]H_o : \mu _1 - \mu_2 = 0[/tex]
The alternative hypothesis is [tex]H_a : \mu _1 - \mu_2 \ne 0[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}[/tex]
[tex]t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}[/tex]
[tex]t = - 0.0899[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = 32+ 32 - 2[/tex]
[tex]df = 62[/tex]
The significance [tex]\alpha[/tex] is evaluated as
[tex]\alpha = (C - 100 )\%[/tex]
=> [tex]\alpha = (95 - 100 )\%[/tex]
=> [tex]\alpha =0.05[/tex]
The critical value is evaluated as
[tex]t_c = 2 * t_{0.05 , 62}[/tex]
From the student t- distribution table
[tex]t_{0.05, 62} = 1.67[/tex]
So
[tex]t_c = 2 * 1.67[/tex]
=> [tex]t_c = 3.34[/tex]
given that
[tex]t_c > t[/tex] we fail to reject the null hypothesis so this mean that the result do not differ
