One year, the population of a city was 99,000. Several years later it was 82,170. Find the percent decrease.
1 answer:
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Answer:
a. P(C) = 0.03
P(test+ | C) = 0.89
P(test- | no C) = 0.93
b. Tree diagram is attached.
c. P(test+) = 0.0946
d. P(C | test+) = 0.2822
Step-by-step explanation:
a. Let C denote Lung Cancer The probability that a heavy smoker will develop lung cancer is 3% i.e. 0.03.
If lung cancer is present, the CT scan result comes out to be positive 89% of the times while if lung cancer is not present, the results are negative 93% of the time. Let test+ denote that the test result is positive and test- denote that the test result is negative.
P(C) = 0.03
P(test+ | C) = 0.89
P(test- | no C) = 0.93
b. The first pair of branches represent whether the individual has cancer or not.
P(C) = 0.03
P(no C) = 1 - 0.03 = 0.97
The second pair of branches will represent if the test result came out positive or negative.
For patients with cancer, P(test+) = 0.89, P(test-) = 1 - 0.89 = 0.11
For patients with no cancer, P(test -) = 0.93, P(test+)= 1 - 0.93 = 0.07.
The tree diagram is attached as an image file.
c. P(test+) = P(C)*P(test+| C) + P(No C)*P(test+| No C)
= (0.03)(0.89) + (0.97)(0.07)
= 0.0267 + 0.0679
P(test+) = 0.0946
d. P(C | test+) = P(C ∩ test+) / P(test+)
= P(C)*P(test+ | C) / P(test+)
= (0.03)*(0.89) / 0.0946
P(C | test+) = 0.2822
Answer:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.
Step-by-step explanation:
Data given and notation
represent the mean height for the sample
represent the population standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 200, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since is a one sided test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.