found Bentley my basset hound while meditating…we were sitting on grass together in the sun☀️🐾🌿
Good morning 🚴♀️🧘♀️
recently discovered this album…on repeat
sunrise 🌅
Good morning 🚴♀️🧘♀️
movie night 🍿 
❤️ 
Good morning :) 🧘♀️
morning coffee 
Good morning 🚴♀️🧘♀️☕️
Enjoying exploring tetragrammaton. Currently listening to Rick’s interview with Edward Norton🎧
sunrise 🌅
Good morning 🚴♀️🧘♀️🚶♀️
While meditating last night a memory of a moment in time came back to me that I just can’t believe I had forgotten…a beautiful surprise🧘♀️
Happy to hear it went well tanel :)
Okay so keeping in mind your sample size is 28, benchmark is 14, average time spent with customers is 9.07, std dev is 1.86, std error is 0.35 which leads to a t-stat of -13.99. The hypothesis could be any of following:
P-value
5% critical value (one-sided upper tail test)
1.HO: m <14 (benchmark)
2.HA: m>14 (benchmark)
3. Test using p-value:
T-stat: remember coefficient-benchmark/ std error which is 9.07-14/0.35 =-13.99
P-value: 1.00
4.Decision rule at 5% significance level: reject null if p<0.05:
a)p-value >0.05 fail to reject
5.Conclusion
There is insufficient evidence at the 5% significance level to conclude that the average time spent per customer is greater than 14 minutes.
P-value
5% critical value (one-sided lower tail test)
1.HO: m >14 (benchmark)
2.HA: m<14 (benchmark)
3. Test using p-value:
T-stat: remember coefficient-benchmark/ std error which is 9.07-14/0.35 =-13.99
P-value: 0.00
4.Decision rule at 5% significance level: reject null if p<0.05:
a)p-value <0.05 reject
5.Conclusion
There is sufficient evidence at the 5% significance level to conclude that the average time spent per customer is less than 14 minutes.
P-value benchmark, any difference or not the same
5% critical value (Two-sided test)
1.HO: m =14 (benchmark)
2.HA: m≠ 14 benchmark)
3. Test using p-value:
T-stat: remember coefficient-benchmark/ std error which is 9.07-14/0.35 =-13.99
P-value: 0.00
4.Decision rule at 5% significance level: reject null if p<0.05:
a)p-value <0.05 reject
5.Conclusion
There is sufficient evidence the average time spent per customer is not at benchmark at 5% significance level.
Confidence Interval benchmark (two-sided test)
1. HO: μ = b (14)
2.HA: μ ≠ b (14)
3. Test using 95% confidence interval:
(8.38, 9.76) lower bound and upper bound.
4. Decision rule: reject if 14 outside of the 95% CI
a) 14 is outside the CI - reject
5. Conclusion
We are 95% confident that the average time spent per customer is not performing at the original benchmark of 14 being the time taken before the training was implemented. Instead, the average time post training is now between 8.38 and 9.76.
Confidence Interval benchmark (two-sided test)
1. HO: μ = b (14)
2.HA: μ ≠ b (14)
3. Test using 99% confidence interval:
(8.16, 9.98) lower bound and upper bound
4. Decision rule: reject if 14 outside of the 99% CI
a) 14 is outside the CI - reject
5. Conclusion
We are 99% confident that the average time spent per customer is not performing at the original benchmark of 14 being the time taken before the training was implemented. Instead, the average time post training is now between 8.16 and 9.98.
Hope this helps tanel :)