Answer:
There is no sufficient evidence to support the claim that wedding cost is less than $30000.
Step-by-step explanation:
Values (x) ∑(Xi-X)^2
----------------------------------
29.1 0.1702
28.5 1.0252
28.8 0.5077
29.4 0.0127
29.8 0.0827
29.8 0.0827
30.1 0.3452
30.6 1.1827
----------------------------------------
236.1 3.4088
Mean = 236.1 / 8 = 29.51
![S_{x}=\sqrt{3.4088/(8-1)}=0.6978](https://tex.z-dn.net/?f=S_%7Bx%7D%3D%5Csqrt%7B3.4088%2F%288-1%29%7D%3D0.6978)
Statement of the null hypothesis:
H0: u ≥ 30 the mean wedding cost is not less than $30,000
H1: u < 30 the mean wedding cost is less than $30,000
Test Statistic:
![t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861](https://tex.z-dn.net/?f=t%3D%5Cfrac%7BX-u%7D%7BS%2F%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B29.51-30%7D%7B0.6978%2F%5Csqrt%7B8%7D%7D%3D%20%5Cfrac%7B-0.49%7D%7B0.2467%7D%3D-1.9861)
Test criteria:
SIgnificance level = 0.05
Degrees of freedom = df = n - 1 = 8 - 1 = 7
Reject null hypothesis (H0) if
![t](https://tex.z-dn.net/?f=t%3C-t_%7B0.05%2Cn-1%7D%5C%5C%20t%3C-t_%7B0.05%2C8-1%7D%5C%5C%20t%3C-t_%7B0.05%2C7%7D)
Finding in the t distribution table α=0.05 with df=7, we have
![t_{0.05,7}=2.365](https://tex.z-dn.net/?f=t_%7B0.05%2C7%7D%3D2.365)
= -1.9861 > -2.365
Result: Fail to reject null hypothesis
Conclusion: Do no reject the null hypothesis
u ≥ 30 the mean wedding cost is not less than $30,000
There is no sufficient evidence to support the claim that wedding cost is less than $30000.
Hope this helps!