find the common difference of the arithmetic sequence. 9, 13, 17, 21, . . . a. nine thirteenths b. 4 c. one and four ninths d. 2
1 answer:
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Answer:
a)
The 95% confidence interval would be given by (0.487;0.553)
b) After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
Part a
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and
. And the critical value would be given by:
The confidence interval for the mean is given by the following formula:
If we replace the values obtained we got:
The 95% confidence interval would be given by (0.487;0.553)
Part b
After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes at 5% of significance.