If 10% equals $50 how much is 25%
2 answers:
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Given:
p = 20% = 0.2, sample proportion
n = 5000, sample size.
Confidence level = 95%
The confidence interval is
(p - 1.96k, p + 1.96k)
where
Therefore the 95% confidence interval is
(0.1943, 0.2057) = (19.4%, 20.6%)
Answer: The 95% confidence interval is (19.4%, 20.6%)
p = 20% = 0.2, sample proportion
n = 5000, sample size.
Confidence level = 95%
The confidence interval is
(p - 1.96k, p + 1.96k)
where
Therefore the 95% confidence interval is
(0.1943, 0.2057) = (19.4%, 20.6%)
Answer: The 95% confidence interval is (19.4%, 20.6%)
Answer:
The correct option is (b).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
The confidence interval for population mean can be computed using either the <em>z</em>-interval or <em>t</em>-interval.
The <em>t</em>-interval is used if the following conditions are satisfied:
- The population standard deviation is not known
- The sample size is large enough
- The population from which the sample is selected is normally distributed.
For computing a (1 - <em>α</em>)% confidence interval for population mean , it is necessary for the population to normally distributed if the sample selected is small, i.e.<em>n</em> < 30, because only then the sampling distribution of sample mean will be approximated by the normal distribution.
In this case the sample size is, <em>n</em> = 28 < 30.
Also it is provided that the systolic blood pressure is known to have a skewed distribution.
Since the sample is small and the population is not normally distributed, the sampling distribution of sample mean will not be approximated by the normal distribution.
Thus, no conclusion can be drawn from the 90% confidence interval for the mean systolic blood pressure.
The correct option is (b).