n = 7 x ˉ = ∑ x n = 36 7 = 5.14285714 s 2 = ∑ x 2 − ( ∑ ( x ) ) 2 n n − 1 = 220 − 185.142857 6 = 5.8095 s = s 2 = 5.8095 = 2.4103 n=7\\\bar x={\sum x\over n}={36\over7}=5.14285714\\ s^2={\sum x^2-{(\sum(x))^2\over n}\over n-1}={220-185.142857\over6}=5.8095\\s=\sqrt{s^2}=\sqrt{5.8095}=2.4103 n = 7 x ˉ = n ∑ x = 7 36 = 5.14285714 s 2 = n − 1 ∑ x 2 − n ( ∑ ( x ) ) 2 = 6 220 − 185.142857 = 5.8095 s = s 2 = 5.8095 = 2.4103
The hypotheses are,
H 0 : μ = 4.25 v s H 1 : μ > 4.25 H_0:\mu=4.25\\vs\\H_1:\mu\gt4.25 H 0 : μ = 4.25 v s H 1 : μ > 4.25
The test statistic is,
t = x ˉ − μ s n = 5.14 − 4.25 2.4103 7 = 0.8929 0.911 = 0.98 t={\bar x-\mu\over {s\over\sqrt{n}}}={5.14-4.25\over{2.4103\over\sqrt{7}}}={0.8929\over0.911}=0.98 t = n s x ˉ − μ = 7 2.4103 5.14 − 4.25 = 0.911 0.8929 = 0.98
The critical value is,
t α , n − 1 = t 0.1 , 6 = 1.439756 t_{\alpha,n-1}=t_{0.1,6}=1.439756 t α , n − 1 = t 0.1 , 6 = 1.439756
The null hypothesis is rejected if, t > t 0.1 , 6 t\gt t_{0.1,6} t > t 0.1 , 6
Since t = 0.98 < t 0.1 , 6 = 1.439756 t=0.98\lt t_{0.1,6}=1.439756 t = 0.98 < t 0.1 , 6 = 1.439756 , the null hypothesis is not rejected. Therefore, we can conclude that there is no sufficient evidence to show that the average family size is more than the national average at 1% level of significance.
Comments