Complete Question
From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:
a. A point estimate of [tex]\mu[/tex] and its 90% margin of error.
b. A 95% confidence interval for [tex]\mu[/tex].
Answer:
a
[tex]\= x = 13.4[/tex] . [tex]E = 2.3[/tex]
b
[tex]10.7 < \mu < 16.1 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 18
The 90% confidence interval is (11.1, 15.7)
Generally the point estimate of [tex]\mu[/tex] is mathematically evaluated as
[tex]\= x = \frac{11.1 + 15.7 }{2}[/tex]
=> [tex]\= x = 13.4[/tex]
Generally the margin of error is mathematically evaluated as
[tex]E = \frac{15.7 - 11.1}{2 }[/tex]
=> [tex]E = 2.3[/tex]
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally the equation for the lower limit of the confidence interval is
[tex]\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1[/tex]
=> [tex]13.4 - 0.3877 s = 11.1[/tex]
=> [tex]s = 5.932[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 1.96 * \frac{5.932}{\sqrt{18} }[/tex]
=> [tex]E = 2.7 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex]13.4 - 2.7 < \mu < 13.4 + 2.7 [/tex]
=> [tex]10.7 < \mu < 16.1 [/tex]
