Since we do not know σ, however, a t interval is more appropriate. The TI-83/84 calculator returns a
t -interval of (71.085, 75.915). Here, t * = 2.4049, so that the resulting interval is slightly wider, as
expected, than the interval obtained by assuming an approximately normal distribution.
The problem states that we have an SRS and n = 49. Since we do not know s , but have a large
sample size, we are justified in using t procedures.
(a) C = 0.95 t = 2.021 (from Table B with df = 40; from a TI-84 with the invT function, the
exact value of t = 2.0106 with df = 48). Thus, . Using the
TI-83/84, we get (186.28, 189.72) based on 48 degrees of freedom. Since 190 is not in this
interval, it is not a likely population value from which we would have gotten this interval. There
is some doubt about the company’s claim.
(b) Let μ = the true mean operating time of the company’s pens.
• H 0 : μ = 190.• H (^) A : μ ≠ 190.
(The wording of the questions tells us H (^) A is two sided.)
• We will use a one-sample t -test. Justify the conditions needed to use this procedure.
• Determine the test statistic (t ) and use it to identify the P -value.
• Compare the P -value with α. Give a conclusion in the context of the problem.
It is not appropriate because confidence intervals use sample data to make estimates about unknown
population values. In this case, the actual difference in the ages of actors and actresses is known, and
the true difference can be calculated.
Solutions to Cumulative Review Problems
Let X = the number of heads. Then X has B (1000, 0.5) because the coin is fair. This binomial can be
approximated by a normal distribution with mean = 1000(0.5) = 500 and standard deviation
calculator, normalcdf(–1.9,1.9 ).
μ (^) x = 0(0.3) + 1(0.37) + 2(0.46) + 3(0.10) + 4(0.04) = 1.75.
(a) P (draw a red marble) = 4/9
(b) Average wait