I have attached the data from the excel spreadsheet. 1 - 31. please show all work.
a) Null hypothesis H0:
minutes
Alternative Hypothesis Ha:
Minutes
alpha=0.05
t= 175.27-144/139.84/sqrt(30)
t= 31.27/139.84/5.48
t= 31.27/25.52
t= 1.23
degrees of freedom= n-1=30-1=29
The p-value is .228582.The result is not significant because p < .05.
Decision: REJECT NULL HYPOTHESIS H0.
Conclusion: We have sufficient evidence to conclude that the population mean is different from 144 minutes.
b) i) random sampling
ii) normality of data distribution
iii) population variance unknown.
c) For normality we can draw Q-Q plot Histogram.
d)
Since data is not normally distributed but data is large enough and t test robust to normality condition therefore it is valid.
I have attached the data from the excel spreadsheet. 1 - 31. please show all work....
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