 # I have a question below revolving around two tailed tests, confidence intervals, etc, and I'm really woowheedr 2022-07-01 Answered
I have a question below revolving around two tailed tests, confidence intervals, etc, and I'm really struggling to figure out how to find the correct answers. It's 3am and this is the last question for an assignment due later in the day, and for some reason I can't seem to properly wrap my mind around t-tests and two tail tests and I'm slowly losing my mind. This is my last ditch-attempt and I would really, really appreciate some help.
Below you'll find the question and my attempt at some of the answers, which are probably way off.
Monthly profits (in million dollars) of two phone companies were collected from the past five years (60 months) and some statistics are given in the following table.
Company | Mean | SD
Rogers | 123.7|25.5
Bell | 242.7|15.4
(a) Find a point estimate for the difference in the average monthly profits of these two phone companies.
I subtracted the two means (242.7-123.7) and got 119.
(b) What is the margin of error for a 99% confidence interval for the difference in the average monthly profits of these two phone companies?
z = (1 - 0.99)/2 z = 0.01/2 z =0.005 1 - 0.005 = 0.995 p(z < 3.275) = 0.995
From looking online, I found this formula but I couldn't figure out how to fill in the values: ME = t(stdev/√n)
I don't know where to go from here. I know the formulas for the confidence intervals, but I don't understand what I'm supposed to do given that there are two standard devs and means? I don't know whether n is 60 or 2.
(c) Based on the constructed 99% confidence interval in part (b), can we conclude that there is a difference in the average monthly profits of these two phone companies?
(d) Given the level of significance α = 1%, test whether Bell's average monthly profit is more than Rogers'.
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a) It is correct:${\mu }_{B}-{\mu }_{R}={\overline{x}}_{B}-{\overline{x}}_{R}=242.7-123.7=119.$
b) Confidence interval for the difference of two population means:${\mu }_{B}-{\mu }_{R}=\left({\overline{x}}_{B}-{\overline{x}}_{R}\right)±\underset{\text{ME}}{\underset{⏟}{{t}_{\alpha /2}\cdot \underset{\text{SE}}{\underset{⏟}{\sqrt{\frac{{s}_{B}^{2}}{{n}_{1}}+\frac{{s}_{R}^{2}}{{n}_{2}}}}}}}\phantom{\rule{0ex}{0ex}}{t}_{0.01/2;df=60-1}={t}_{0.005,59}=2.6618;\phantom{\rule{0ex}{0ex}}\text{ME}=2.6618\cdot \sqrt{\frac{{15.4}^{2}}{60}+\frac{{25.5}^{2}}{60}}=10.24;\phantom{\rule{0ex}{0ex}}{\mu }_{B}-{\mu }_{R}=\left(242.7-123.7\right)±10.24=\left(108.76,129.24\right)$
Note: ME − margin of error, SE − standard error.
c) Yes, because: ${\mu }_{B}-{\mu }_{R}>0$
d)
There is a significant evidence to conclude that the Bell's average monthly profit is more than Rogers'.

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