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.
Answer:
0.142
Step-by-step explanation:
divide -1 by 7 = -0.142
opposite of that would be positive
Answer: 3 hours
Step-by-step explanation:
From the question, we are informed that WHUR Radio uses 1/5 of its of its 10 hours of daytime broadcast for commercial and 1/2 for music.
Hours used for commercial:
= 1/5 × 10
= 2 hours
Hours used for music:
= 1/2 × 10
= 5 hours
Hours left for news will be:
= 10 hours - 2 hours - 5 hours
= 3 hours
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).