Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longboard​ (L) users and 30 shortboard​ (S) users. Treat these data as though they were from two independent random samples. Test the hypothesis that the mean days surfed for all longboarders is larger than the mean days surfed for all shortboarders​ (because longboards can go out in many different surfing​ conditions). Use a level of significance of 0.05.
Longboard :4,8,9,4,9,7,9,6,6,1 1,1 3, 12, 10, 151316,12,10. 12,18,20.15,10.15,19,21,9,22, 1923 Shortboard: 6,4,4,6,7,7,7,10,4,6,7,5,8,9,4,15,13,9,12,11,12,13,9,11,13,15,9,19,20,11
Determine the hypotheses for this test. Choose the correct answer below. Find the test statistic for this test. Find the p-value for this test. What is the conclusion for this test? Reject H0. The mean days surfed for longboarders is not significantly larger than the mean days surfed for all shortboarders Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders Do not reject H0. The mean days surfed for longboarders is not significantly larger than the mean days surfed for all shortboarders

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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