Answer:
Step-by-step explanation:
The mean of the set of data given is
Mean = (29.7 + 29.4 + 31.7 + 29.0 + 29.1 + 30.5 + 29.1 + 29.8)/8 = 29.788 = $29788
Standard deviation = √(summation(x - mean)/n
n = 8
Summation(x - mean) = (29700 - 29788)^2 + (29400 - 29788)^2 + (31700 - 29788)^2 + (29000 - 29788)^2 + (29100 - 29788)^2 + (30500 - 29788)^2 + (29100 - 29788)^2 + (29800 - 29788)^2 = 5888752
Standard deviation = √(5888752/8) = 857.96
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ < 30000
For the alternative hypothesis,
µ > 30000
It is a right tailed test
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 8,
Degrees of freedom, df = n - 1 = 8 - 1 = 7
t = (x - µ)/(s/√n)
Where
x = sample mean = 29788
µ = population mean = 30000
s = samples standard deviation = 857.96
t = (29788 - 30000)/(857.96/√8) = - 0.7
We would determine the p value using the t test calculator. It becomes
p = 0.253
Since alpha, 0.05 < than the p value, 0.253, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, the sample data did not show evidence that the mean wedding cost is not less than $30,000 as advertised
