Download Friday, October 23 rd

Homework #3
1) According to SPSS:
Mean
s
SE
=
=
=
3.54
6.11
0.54
Technically, you should use t for this calculation because σ is unknown. For df, the
closest you have in the table is 120, so go ahead and use that…
95% CI
=
Mean  t/2 * (s / sqrt(n))
=
3.54  1.98 * (6.11 / sqrt(130))
=
3.54  1.98 * 0.54
=
3.54  1.06
95% CI
=
[2.48 – 4.60]
But if you used z…
95% CI
=
Mean  z/2 * (s / sqrt(n))
=
3.54  1.96 * (6.11 / sqrt(130))
=
3.54  1.96 * 0.54
=
3.54  1.05
95% CI
=
[ 2.49– 4.59]
Note that the values for the two methods are so close because the critical values are so
close because the degrees of freedom are so high.
SPSS output below. The CI is a little different due to differences in rounding and
because SPSS uses t rather than z and uses the t for 129 degrees of freedom rather than
120 because it can calculate the correct value for t which does not appear in our table.
One-Sample Test
Test Value = 0
Skip_Hrs
t
6.611
df
129
Sig. (2tailed)
.000
Mean
Difference
3.5423
95% Confidence
Interval of the
Difference
Lower
2.482
Upper
4.602
2) The data from HW 1 Q3): 4.5, 6, 6, 7, 7, 7, 8
Mean
=
6.5
s
=
1.12
SE
=
1.12 / (sqrt(7))
Plug these values into the formula for the CI
90% CI
=
Mean  t/2 * (s / sqrt(n))
=
6.5  1.943 * (.42)
=
6.5  .72
90% CI
=
[5.68 – 7.32]
=
.42
=
Mean  t/2 * (s / sqrt(n))
=
6.5  2.447 * (.42)
=
6.5  1.03
95% CI
=
[5.47 – 7.53]
Notice that the 95% CI is wider than the 90% CI.
95% CI
3) Here you would have to use z rather than t even though we don’t technically know σ;
not ideal. If I ask you to do this on an exam, I’ll be sure to provide you with σ.
n
=
[z/2 * s / E]2
=
[1.96 * 6.11 / 1]2
=
[11.98]2
=
143.42
Round up to 144
4) Fill in the blanks: According to the central limit theorem, the sampling distrubtion of a
population will be approximately normal if n is sufficiently large (n > 30). Also,
the population parameter should be equal to the mean of the sampling distribution
and the standard error will be equal to s / sqrt (n).
5) A variety of answers would suffice, including the following:
 To ensure an unbiased estimator (i.e., random sample).
 To decrease the variability of our estimator (i.e., increase its reliability).
 To enable us to use the Central Limit Theorem as a way of modeling chance
variation in our sample.
 To increase the likelihood of obtaining a representative sample.