Fifty randomly selected individuals were timed completing a tax form. the sample mean was 23.6 minutes; the sample standard devi
1 answer:
Given:
n = 50, sample size
, sample mean
s = 2.4 min, sample standard deviation.
The confidence interval is
At the 99% confidence level, t* (from the student's t-distribution) is
t* = 2.68
Therefore
t*(s/√n) = 2.68*(2.4/√50) = 0.9096
The confidence interval is
(23.6-0.9096, 23.6+0.9096) ≈ (22.69, 24.51)
Answer: (22.7, 24.5)
n = 50, sample size
s = 2.4 min, sample standard deviation.
The confidence interval is
At the 99% confidence level, t* (from the student's t-distribution) is
t* = 2.68
Therefore
t*(s/√n) = 2.68*(2.4/√50) = 0.9096
The confidence interval is
(23.6-0.9096, 23.6+0.9096) ≈ (22.69, 24.51)
Answer: (22.7, 24.5)
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Step-by-step explanation:
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Answer:
Assuming a standard significance level of the best conclusion for this case would be:
4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).
Because
If we select a significance level lower than 0.034 then the conclusion would change.
Step-by-step explanation:
Data given
n=300 represent the random sample taken
X=120 represent the people who have a smart phone
estimated proportion of people who have a smart phone
is the value that we want to test
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to test the claim that the true proportion of people who have a smart phone is higher than 0.35, the system of hypothesis are.:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
Statistical decision
Since is a right tailed test the p value would be:
Assuming a standard significance level of the best conclusion for this case would be:
4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).
Because
If we select a significance level lower than 0.034 then the conclusion would change.