AB intersects CD at point E. What is the value of X in degrees?
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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).
Draw a diagram to illustrate the problem as shown in the figure below.
The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.
The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m
The model for tide depth is
That is,
d = -1.5 cos(0.5027t) + 4
where
d = depth, m
t = time, hours
A plot of the function confirms that the model is correct.
The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.
The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m
The model for tide depth is
That is,
d = -1.5 cos(0.5027t) + 4
where
d = depth, m
t = time, hours
A plot of the function confirms that the model is correct.