Answer:
Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders
Step-by-step explanation:
The null hypothesis is that the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders
H0: μL > μs against the claim Ha: μL≤ μs
the alternate hypothesis is the mean days surfed for all long boarders isless or equal to the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)
The test statistic is
t= x1- x2/ √s1/n1+ s2/n2
1) Calculations
Longboards
Mean
ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30
=377/30
=12.5667
Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1
=[831-(-13)²/30]/29
=831-5.6333/29
=825.3667/29
=28.4609
Shortboard Mean
ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30
=288/30
=9.6
Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1
=[ 3270-(288)2/30]/29
=3270-2764.8/29
=505.2/29
=17.4207
2) Putting values in the test statistic
t=|x1-x2|/√S²1/n1+S²2/n2
t =|12.5667-9.6|/√28.4609/30+17.4207/30
t =|2.9667|/√0.9487+0.5807
t=|2.9667|/√1.5294
t=|2.9667|/1.2367
t=2.3989
3) Degree of freedom =n1+n2-2=30+30-2=58
4) The critical region is t ≤ t(0.05) (58) =1.6716
5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58) ≤ 1.6716 we do not reject H0.
The p-value is 0.008969. The result is significant at p <0 .05.
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