Respuesta :
Answer:
a) From definition the proportion of successes would be:
[tex] \hat p = \frac{x}{n} = \frac{7}{7+3}= 0.7[/tex]
b) Data : 1,1,0,0,1,1,1,0,1,1
x = 7 for this case
[tex] \hat p = \frac{7}{10}=0.7[/tex]
The proportion in part (a) is exactly equal to the mean in part (b)
c) We have actually 7 successes and 3 fails and we want this condition:
[tex] \frac{x}{n}=0.6[/tex]
The new value of n would be 25 so then we can solve for x and we got:
[tex] x = 25*0.6= 15[/tex]
So we need 15 successes and we have 7 so then we need [tex] 15-7 = 8[/tex] successes in the new 15 cars in order to satisfy the condition.
Step-by-step explanation:
Data: S S F F S S S F S S
S means success and F damage. We have 7 successes and 3 damages for this case
(a) What is the value of the sample proportion of successes x/n?(
From definition the proportion of successes would be:
[tex] \hat p = \frac{x}{n} = \frac{7}{7+3}= 0.7[/tex]
(b) Replace each S with a 1 and each F with a 0. Then calculate x for this numerically coded sample.How does x compare with x/n?
Data : 1,1,0,0,1,1,1,0,1,1
x = 7 for this case
[tex] \hat p = \frac{7}{10}=0.7[/tex]
The proportion in part (a) is exactly equal to the mean in part (b)
(c) Suppose it is decided to include 15 more cars in the experiment. How many of these would have to be S's to give x/n = 0.6 for the entire sample of 25 cars?
We have actually 7 successes and 3 fails and we want this condition:
[tex] \frac{x}{n}=0.6[/tex]
The new value of n would be 25 so then we can solve for x and we got:
[tex] x = 25*0.6= 15[/tex]
So we need 15 successes and we have 7 so then we need [tex] 15-7 = 8[/tex] successes in the new 15 cars in order to satisfy the condition.
