1. Refer to the PairedTTest tab in the assignment spreadsheet. Note that this co

Q: 1. Refer to the PairedTTest tab in the assignment spreadsheet. Note that this co

In this case. t stat = -0.13706 which is lesser than 2.010635(t critical) Also we have p value = 0.8916 > 0.05 So fail to reject H0 i.e Dia1 had not lower mean pressure than Dia2. Conclusions would be False : Dia1 had a lower mean pressure than Dia2 but the measurements were much more spread out from the mean False. We reject Ho since p = 0.199130314

ID: 2906613 • Letter: 1

Question

1. Refer to the PairedTTest tab in the assignment spreadsheet. Note that this compares Dia 1 (before) and Dia2 (after) measurements from the same group 2.

True/False. Dia1 had a lower mean pressure than Dia2 but the measurements were much more spread out from the mean. _______ (1 pt)

True/False. We reject Ho since p = 0.199130314

Wt_Grp_1 Wt_Grp_2

Mean 182.3478 183.1111

Variance 350.1462 414.8718

Observations 23 27

Pooled Variance 385.2059

Hypothesized Mean Difference. 0

df 48

t Stat -0.13706

P(T<=t) two-tail 0.891559

t Critical two-tail 2.010635

Explanation / Answer

In this case.

t stat = -0.13706 which is lesser than 2.010635(t critical)

Also we have p value = 0.8916 > 0.05

So fail to reject H0

i.e Dia1 had not lower mean pressure than Dia2.

Conclusions would be

False : Dia1 had a lower mean pressure than Dia2 but the measurements were much more spread out from the mean

False. We reject Ho since p = 0.199130314

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