QUESTION IMAGE
Question
a simple random sample of size n is drawn from a population that is normally distributed. the sample mean, \\( \bar { x } \\) is found to be 106, and the sample standard deviation, s, is found to be 10
(a) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 23
(b) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 19
(c) construct a 96% confidence interval about \\( \mu \\) if the sample size, n, is 23
(d) could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?
click the icon to view the table of areas under the t - distribution.
(b) construct a 98% confidence interval about \\( \mu \\) if the sample size, n, is 19.
lower bound: 100.1, upper bound: 111.9
(use ascending order. round to one decimal place as needed.)
how does decreasing the sample size affect the margin of error, e?
a. as the sample size decreases, the margin of error decreases.
b. as the sample size decreases, the margin of error stays the same.
c. as the sample size decreases, the margin of error increases.
(c) construct a 96% confidence interval about \\( \mu \\) if the sample size, n, is 23
lower bound: \\( \square \\), upper bound: \\( \square \\)
(use ascending order. round to one decimal place as needed.)
Step1: Find the critical value
For a 96% confidence interval with \(n = 23\) (degrees of freedom \(df=n - 1=23 - 1 = 22\)), the significance level \(\alpha=1 - 0.96=0.04\), and \(\frac{\alpha}{2}=0.02\).
Using the t - distribution table, \(t_{\frac{\alpha}{2},df}=t_{0.02,22}\approx 2.189\).
Step2: Calculate the margin of error
The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 10\), \(n = 23\).
\(E=2.189\times\frac{10}{\sqrt{23}}\approx2.189\times\frac{10}{4.796}\approx2.189\times2.085\approx4.56\).
Step3: Calculate the confidence interval
The sample mean \(\bar{x}=106\).
The lower bound \(=\bar{x}-E=106 - 4.56 = 101.4\).
The upper bound \(=\bar{x}+E=106+4.56 = 110.6\).
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Lower bound: \(101.4\), Upper bound: \(110.6\)