Answer:
a
The null hypothesis is [tex]H_o : \sigma^2_1 = \sigma^2 _2[/tex]
The alternative hypothesis is [tex]H_a : \sigma_1 ^2 > \sigma^2_2[/tex]
b
[tex]F_{critical} = 1.8608[/tex]
c
[tex]F = 2.9085[/tex]
d
The decision rule is
Reject the null hypothesis
e
There is sufficient evidence to support the researchers claim
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 30[/tex]
The sample variance for elementary school is [tex]s^2_1 = 8324[/tex]
The second sample size is [tex]n_2 = 30[/tex]
The sample variance for the secondary school is [tex]s^2_2 = 2862[/tex]
The significance level is [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o : \sigma^2_1 = \sigma^2 _2[/tex]
The alternative hypothesis is [tex]H_a : \sigma_1 ^2 > \sigma^2_2[/tex]
Generally from the F statistics table the critical value of [tex]\alpha = 0.05[/tex] at first and second degree of freedom [tex]df_1 = n_1 - 1 = 30 - 1 = 29[/tex] and [tex]df_2 = n_2 - 1 = 30 - 1 = 29[/tex] is
[tex]F_{critical} = 1.8608[/tex]
Generally the test statistics is mathematically represented as
[tex]F = \frac{s_1^2 }{s_2^2}[/tex]
=> [tex]F = \frac{8324 }{2862}[/tex]
=> [tex]F = 2.9085[/tex]
Generally from the value obtained we see that [tex]F > F_{critical }[/tex] Hence
The decision rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to support the researchers claim
