Betty has several of the standard six-sided dice that are common in many board games. If Betty rolls one of these dice, what is
1 answer:
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Answer:
a)
b)
If we compare the p value and the significance level given we see that we have enough evidence to reject the null hypothesis at 5% of significance.
Step-by-step explanation:
Data given and notation
n=114 represent the random sample taken
estimated proportion of people that their approval rating might have changed
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Hypothesis
We need to conduct a hypothesis in order to test the claim that true proportion of people that their approval rating might have changed is 0.58 or no.:
Null hypothesis:
Alternative hypothesis:
Part a
(1)
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
Part b: Statistical decision
The significance level provided . The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
If we compare the p value and the significance level given we see that we have enough evidence to reject the null hypothesis at 5% of significance.
Answer:
a) P(t>2.120) = 0.025
b) P(t<1.337) = 0.90
c) P(t<-1.746) = 0.05
d) P(t>2.583) = 0.01
e) P(-2.120<t<2.120) = 0.95
f) P(-1.746<t<1.746) = 0.90
Step-by-step explanation:
The question is incomplete:
For a t distribution with 16 degrees of freedom, find the area, or probability, in each region.
a. To the right of 2.120. (Use 3 decimals.)
b. To the left of 1.337. (Use 2 decimals.)
c. To the left of -1.746. (Use 2 decimals.)
d. To the right of 2.583. (Use 2 decimals.)
e. Between -2.120 and 2.120. (Use 2 decimals.)
f. Between -1.746 and 1.746. (Use 2 decimals.)
We have a t-distribution with 16 degrees of freedom.
We can calculate the probabilities with a spreadsheet, a table or an applet.
We have to calculate the probabilities of:
a) P(t>2.120) = 0.025
b) P(t<1.337) = 0.90
c) P(t<-1.746) = 0.05
d) P(t>2.583) = 0.01
e) P(-2.120<t<2.120) = 0.95
f) P(-1.746<t<1.746) = 0.90
