Examine by a test

[2suchy7]

XYZ has counted the number of customers on 5 weekdays:

Mondays Tuesdays Wednesdays Thursdays Fridays
45 39 67 53 35
47 41 69 57 37
49 43 45 59 51
31 49 67 61 55
49 55 61 63 65

(Numbers should be in table)

a) Examine by a test if the average number of customers is identical on all days
b) Examine if the assumptions of the above test are met

Could someone show me how to do this sep by step?

 

[galactus]

This appears to be an ANOVA test. Do you have Excel, calculator, or some other software?.

 

[2suchy7]

Yes, I will be doing it in excel and I will be using also Hypostat program, so if you could explain me how can i do this I will be glad.

 

[galactus]

Type your data into Excel in columns A through E. Let the first row be the headings for the days of the week.

*Go to Tools, Data Analysis, ANOVA-One Factor. It is self-explanatory. Make sure you check the "Label in First Row"

After Excel gives you the analysis, look at the p-value, and F vs F-crit.

These tell you rather or not to reject H0 and rather or not the means are equal.

You did not state an alpha level, so I used .05. That is most common.

\(\displaystyle H_{0}: \;\ {\mu}_{1}={\mu}_{2}=......={\mu}^{5}\)

\(\displaystyle H_{a}: \;\ \text{At least one mean is significantly different}\)

If H0 is rejected then they are different. If it is not rejected, then they are the same.

*I still have Excel 2003. You may have to load the Data Analysis pack.

 

[2suchy7]

Ok thanks.
I need to know one more thing. When H0 is rejected?

 

[Deleted member 4993]

Ok thanks.
I need to know one more thing. When H0 is rejected?

If you are asking - Under what condition H[sub:266kdolk]o[/sub:266kdolk] is rejected - the answer is

H[sub:266kdolk]o[/sub:266kdolk] is rejected - when the alternative hypothesis H[sub:266kdolk]a[/sub:266kdolk] cannot be rejected at the given significance level.