# Q2. You collected 500 weeks of data (2500 days total). Based on that you find Tuesday's mean return is 12 bps. Mean return of all days is 2 bps. Stdev across all days is 100 bps. There is no noticeable difference b/w Tuesday stdev vs other weekdays' stdev.Q2d. Based on 95% CI, should you reject or accept that Tuesday's mean return is not different from other weekdays?

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Q2. You collected 500 weeks of data (2500 days total). Based on that you find Tuesday's mean return is 12 bps. Mean return of all days is 2 bps. Stdev across all days is 100 bps. There is no noticeable difference b/w Tuesday stdev vs other weekdays' stdev.
Q2d. Based on 95% CI, should you reject or accept that Tuesday's mean return is not different from other weekdays?

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Step 1

There are total 500 weeks that is total 2,500 days. Therefore, total 500 Tuesdays are calculated. The mean return of Tuesdays is 12 bps. The sample standard deviation for all days is 100 bps.

Here, the population standard deviation is not known, but the sample sizes are very large, therefore, for finding the confidence interval for the mean difference between the Tuesday’s mean return and the all days’ mean return the z-confidence interval can be used.

Let μ1 be the true mean of Tuesday’s.

Let μ2 be the true mean of other week days’.

There are total 500 weeks. Therefore, total 500 Tuesdays are calculated.

The mean return of Tuesdays is 12 bps.

Hence, total return for all Tuesdays are,

Step 2

Mean return of all days is 2 bps. Total number of days is 2,500.

Hence, total return for all days are,

Step 3

Therefore, total return for all w...

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