Transcript ECON103 Tutorial Week 19

ECON103
Week 19
Rob Pryce
www.robpryce.co.uk/teaching
PE7.2 Q3
3. In a random sample of 220 customers who called into a hardware shop, 88 made no purchases, 33
made purchases €100 or less, the remainder made purchases exceeding €100.
(a)
Calculate a 95% confidence interval for the proportion of customers that
(i) made no purchases;
𝐶𝐼 = 𝑝 ± 𝑍𝛼
2
𝑝=
𝜎2
𝑛
88
; 𝑍 = 1.96 ; 𝑛 = 220 ; 𝜎 2 = 𝑝 1 − 𝑝
220
𝐶𝐼 = 0.4 ± 1.96 0.033
𝐶𝐼 = 0.4 ± 0.0647
(0.335, 0.465)
(b) (i) Does it support the claim that 50% of customers made no purchase? No, 0.5 does not lie in CI
PE 7.2 Q5
5. A Health and Safety survey of 200 industrial accidents claims revealed that 53 were
due to untidy working conditions.
(a) Calculate a 90% confidence interval for the proportion of accidents that are due
to untidy working conditions.
𝜎2
𝐶𝐼 = 𝑝 ± 𝑍
𝑛
53
𝑝=
; 𝑍 = 1.645 ; 𝜎 2 = (𝑝 1 − 𝑝 ; 𝑛 = 200
200
𝐶𝐼 = 0.265 ± 0.0513
PE 7.2 Q5
5. A Health and Safety survey of 200 industrial accidents claims revealed that 53 were
due to untidy working conditions.
(a) Calculate the margin of error at 99% confidence.
Standard error
𝐶𝐼 = 𝑝 ±
𝑍
𝜎2
𝑛
𝑀𝐸 = 0.08 (𝑍 = 2.57)
Margin of error
Your Turn - PE7.3 Q7
7. An opinion poll of 410 voters taken a week before an
election showed that 138 intended to vote for the current
party in power (referred to as Party A), 102 for the largest
opposition party (Party B), 67 for others, 82 don’t know,
21 won’t vote at all.
(a) Calculate a 95% confidence interval for the proportion
that will vote for the current party in power. What is
the margin of error for this estimate? Is this party
likely to gain more than 50% of the vote?
(b) Calculate a 95% confidence interval for the proportion
that will vote for the main opposition party.
PE7.4 Q5
5. A survey of regular customers in (i)
restaurants and (ii) pubs were asked whether
they approve of non-smoking in their
restaurant/pub. Their responses are given
below:
Restaurant customers
Pub customers
Approve
64
36
Total surveyed
112
72
PE7.4 Q5
Restaurant customers
Pub customers
Approve
64
36
Total surveyed
112
72
Construct a 95% confidence interval for the difference in the proportions that approve smoking in
restaurant and pubs. What do you conclude?
𝐶𝐼𝑑𝑖𝑓𝑓 = 𝑝1 − 𝑝2
𝜎12 𝜎22
±𝑍
+
𝑛1 𝑛2
𝐶𝐼𝑑𝑖𝑓𝑓 = 0.5714 − 0.5 ± 1.96(0.0752)
𝐶𝐼𝑑𝑖𝑓𝑓 = 0.0714 ± 0.147
𝐶𝐼𝑑𝑖𝑓𝑓 = (−0.07, 0.21)
We conclude there is no significant difference in support for the smoking ban between restaurant
and pub customers (since our confidence interval ranges from negative to positive).
Your Turn - PE7.4 Q7
In a dental practice the number of patients who do not show up for
appointments is thought to be particularly high on Mondays. Form past
records, the number of ‘no shows’ on randomly selected Mondays and
Tuesdays are as follows:
Monday
Tuesday
Number of appointments
248
240
Number of ‘no shows’
52
31
Calculate a 98% confidence interval for the difference in the proportion of ‘no shows’
between Mondays and Tuesdays. Interpret your results.
Questions?
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