Ken normally leaves work at 5:00 pm, but he is leaving 20 minutes late today. He decides to make up time by taking the toll road
1 answer:
You might be interested in
Answer:
Null hypothesis:
Alternative hypothesis:
So the p value is a very low value and using any significance level for example always
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.
Step-by-step explanation:
Data given and notation
represent the number of homeowners who would buy the security system
represent the number of renters who would buy the security system
sample 1
sample 2
represent the proportion of homeowners who would buy the security system
represent the proportion of renters who would buy the security system
z would represent the statistic (variable of interest)
represent the value for the test (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the two proportions differs , the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We need to apply a z test to compare proportions, and the statistic is given by:
(1)
Where
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
Statistical decision
For this case we don't have a significance level provided , but we can calculate the p value for this test.
Since is a two sided test the p value would be:
So the p value is a very low value and using any significance level for example always
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions NOT differs significantly.
Answer:
Option C - Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
Step-by-step explanation:
We are given;
n = 15
t-value = 1.66
Significance level;α = 0.05
So, DF = n - 1 = 15 - 1 = 14
From the one-sample t - table attached, we can see that the p - value of 0.06 at a t-value of 1.66 and a DF of 14
Now, since the P-value is 0.06,it is greater than the significance level of 0.05. Thus we do not reject the null hypothesis. We conclude that there is not sufficient evidence that the true diameter differs from 0.5 in.
