Holly missed 7 of 185 school days this year. Approximately what percent of school days was Holly in attendance?
1 answer:
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Answer:
a) ii. This is a left-tailed test.
b) -1.59
c) -1.301
d) i. reject null hypothesis
e) Option i) The data supports the claim that college students get less sleep than the general population.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 7 hours
Sample mean, = 6.87 hours
Sample size, n = 45
Alpha, α = 0.10
Sample standard deviation, s = 0.55 hours
First, we design the null and the alternate hypothesis
a) We use one-tailed(left) t test to perform this hypothesis.
b) Formula:
Putting all the values, we have
c) Now,
Since,
d) We fail to accept the null hypothesis and reject it.
We accept the alternate hypothesis and conclude that mean number of hours of sleep for all college students is less than 7 hours.
e) Option i) The data supports the claim that college students get less sleep than the general population.
Y =ax² + bx +c
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7