A function is defined by f(x)=5(2-x). What is f(–1)?
1 answer:
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18/24 can be written as 3/4 in simplest form.
Answer:
a) The 90% confidence interval would be given (0.561;0.719).
b)
c) Using the significance level assumed we see that
so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.
Step-by-step explanation:
1) Data given and notation
n=100 represent the random sample taken
X=64 represent were in favor of firing the coach
estimated proportion for were in favor of firing the coach
is the value that we want to test since the problem says majority
represent the significance level (no given, but is assumed)
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
p= population proportion of Americans for were in favor of firing the coach
Part a
The confidence interval would be given by this formula
For the 90% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 90% confidence interval would be given (0.561;0.719).
Part b
We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :
Null Hypothesis:
Alternative Hypothesis:
We assume that the proportion follows a normal distribution.
This is a one tail upper test for the proportion of union membership.
The One-Sample Proportion Test is "used to assess whether a population proportion is significantly (different,higher or less) from a hypothesized value
".
Check for the assumptions that he sample must satisfy in order to apply the test
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
Calculate the statistic
The statistic is calculated with the following formula:
On this case the value of is the value that we are testing and n = 100.
The p value for the test would be:
Part c
Using the significance level assumed we see that
so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.