Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu μ. Assume that the population has a normal distribution. A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s equals =17.6 milligrams. Construct a​ 95% confidence interval for the true mean cholesterol content of all such eggs
A. 175.9 mg less than < mu μ less than <194.1 mg
B. 173.9 mg less than < mu μ less than <196.1 mg
C. 173.8 mg less than < mu μ less than <196.2 mg
D. 173.7 mg less than < mu μ less than <196.3 mg

Answer 1

Answer:

option (C) 173.8 mg less than < mu μ less than <196.2 mg

Step-by-step explanation:

Data provided ;

number of sample, n = 12

Mean = 185 milligram

standard deviation, s = 17.6 milligrams

confidence level = 95%

α = 0.05                                     [for 95% confidence level]

df = n - 1 = 12 - 1 = 11

Now,

Confidence interval = Mean ± E

here,

E is the margin of error = $t_{\alpha/2, df}\frac{s}{\sqrt{n}}$

also,

$t_{\alpha/2, df}$

=  $t_{0.05/2, (11)}$  

=   2.201                         [ from standard t value table]

Thus,

E = $2.201\times\frac{17.6}{\sqrt{12}}$

or

E = 11.182 milligrams ≈ 11.2 milligrams

Therefore,

Confidence interval:

Mean - E < μ < Mean + E

or

185 - 11.2 < μ < 185 + 11.2

or

173.8 < μ < 196.2

Hence,

the correct answer is option (C) 173.8 mg less than < mu μ less than <196.2 mg