Which of these is equivalent to 3^2 ? Select all that apply
1 answer:
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Answer:
Given that is not the easiest equation to solve, and that
the values at the convenient ± 1 sigma,± 2 sigma,± 3 sigma are irrational
using the "rule-of-thumb" "empirical rule" is typically "close enough"
given that the whole probability analysis by definition has some uncertainty in it
Step-by-step explanation:
Answer:
(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).
(b) The confidence level is 68%.
(c) The necessary assumption is that the population is normally distributed.
(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).
Step-by-step explanation:
Let <em>X</em> = temperature of a certain solution.
The estimated mean temperature is, .
The estimated standard deviation is, .
(a)
The general form of a (1 - <em>α</em>)% confidence interval is:
Here,
SS = sample statistic
CV = critical value
SD = standard deviation
It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.
Since the sample size is large use a <em>z</em>-confidence interval.
The critical value of <em>z</em> for 95% confidence interval is:
Compute the confidence interval as follows:
Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).
(b)
The confidence interval is, 37 ± 0.1°C.
Comparing the confidence interval with the general form:
The critical value is,
CV = 1
Compute the value of P (-1 < Z < 1) as follows:
The percentage of <em>z</em>-distribution between -1 and 1 is, 68%.
Thus, the confidence level is 68%.
(c)
The confidence interval for population mean can be constructed using either the <em>z</em>-interval or <em>t</em>-interval.
If the sample selected is small and the standard deviation is estimated from the sample, then a <em>t</em>-interval will be used to construct the confidence interval.
But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.
Thus, the necessary assumption is that the population is normally distributed.
(d)
For <em>n</em> = 10 compute a 95% confidence interval for the temperature as follows:
The (1 - <em>α</em>)% <em>t</em>-confidence interval is:
The critical value of <em>t</em> is:
*Use a <em>t</em>-table for the critical value.
The 95% confidence interval is:
Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).
