Multiply the fraction by 100%.
(2,3) I think. I'm not in Algebra 1, so yeah
Answer:
confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.
(0.79 , 0.89)
Step-by-step explanation:
step 1:-
Given sample survey former UF students n = 1532
84% said they were still in love with Tim Tebow
p = 0.84
The survey sampling error
Given standard error of proportion = 2% =0.02
<u>Step 2</u>:-
The 99% of z- interval is 2.57
The 99% of confidence intervals are
p ± zₐ S.E (since sampling error of proportion = 
on simplification , we get
(0.84 - 0.0514 , 0.84 + 0.0514)
(0.79 , 0.89)
<u>conclusion</u>:-
confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.
(0.79 , 0.89)
Answer:
the first answer
Step-by-step explanation:
1/8 of the absolute value is 1/8 positive
anything from the absolute value is positive because absolute value is the value of how far away the number is from zero.
Answer:
Step-by-step explanation:
1. Null hypothesis: u <= 0.784
Alternative hypothesis: u > 0.784
2. Find the test statistics: z using the one sample proportion test. First we have to find the standard deviation
Using the formula
sd = √[{P (1-P)}/n]
Where P = 0.84 and n = 750
sd =√[{0.84( 1- 0.84)/750]}
sd=√(0.84 (0.16) /750)
SD =√(0.1344/750)
sd = √0.0001792
sd = 0.013
Then using this we can find z
z = (p - P) / sd
z = (0.84-0.784) / 0.013
z =(0.056/0.013)
z = 4.3077
3. Find the p value and use it to make conclusions...
The p value at 0.02 level of significance for a one tailed test with 4.3077 as z score and using a p value calculator is 0.000008254.
4. Conclusions: the results is significant at 0.02 level of significance suck that we can conclude that its on-time arrival rate is now higher than 78.4%.
