Answer:
The upper limit is
[tex]k = 52.94[/tex]
Step-by-step explanation:
From the question we told that
The sample size is [tex]n = 16[/tex]
The sample mean is [tex]\= x = 50[/tex]
The sample variance is [tex]\sigma ^2 = 36[/tex]
For a 95% confidence interval the confidence level is 95%
Given that the confidence level is 95% then the level of significance is mathematically evaluated as
[tex]\alpha = 100 - 95[/tex]
[tex]\alpha = 5 \%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table(reference- math dot armstrong dot edu), the value 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]
Here [tex]\sigma[/tex] is the standard deviation which is mathematically evaluated as
[tex]\sigma = \sqrt{\sigma^2}[/tex]
substituting values
[tex]\sigma = \sqrt{36}[/tex]
=> [tex]\sigma = 6[/tex]
So
[tex]E = 1.96 * \frac{6}{\sqrt{16} }[/tex]
[tex]E = 2.94[/tex]
The 95% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]50 -2.94 < \mu <50 +2.94[/tex]
[tex]47.06 < \mu <52.94[/tex]
The upper limit is
[tex]k = 52.94[/tex]
