Answer:
Explanation:
n = 12
The mean of the set of data given is
Mean, x = (286 + 290 + 285 + 291 + 287 + 275 + 262 + 289 + 274 + 276 + 297 + 262)/12 = 281.2
Standard deviation = √(summation(x - mean)²/n
Summation(x - mean)² = (286 - 281.2)^2 + (290 - 281.2)^2 + (285 - 281.2)^2 + (291 - 281.2)^2 + (287 - 281.2)^2 + (275 - 281.2)^2 + (262 - 281.2)^2 + (289 - 281.2)^2 + (274 - 281.2)^2 + (276 - 281.2)^2 + (297 - 281.2)^2 + (262 - 281.2)^2 = 1409.68
Standard deviation, s = √(1409.68/12) = 10.8
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
1) For the null hypothesis,
H0: µ = 290
For the alternative hypothesis,
H1: µ < 290
This is a left tailed test.
2) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 12,
Degrees of freedom, df = n - 1 = 12 - 1 = 11
t = (x - µ)/(s/√n)
Where
x = sample mean = 281.2
µ = population mean = 290
s = samples standard deviation = 10.8
t = (281.2 - 290)/(10.8/√12) = - 2.82
Test statistic = - 2.82
3) The decision rule is to reject the null hypothesis if the significance level is greater than the p value.
4) We would determine the p value using the t test calculator. It becomes
p value = 0.0083
5) Since alpha, 0.01 > than the p value, 0.0083, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the mean fare has decreased.