Ch. 8 Estimating with Confidence Ch. 8-3 Estimating a Population Mean
a mean Sample:
We want to estimate the true mean final exam score, 𝜇, for a class of AP Stats student at a 96% confidence level. one sample z interval for mean Random: The sample is an SRS. Normal: The population distribution is normal, so the sampling distribution of 𝑥 is approximately normal. Independent: Sampling without replacement so check 10% condition There are at least 10(4) = 40 AP Stats students.
Estimate ± Margin of Error 𝜎 𝑛 𝑥 ± 𝑧 ∗ .96 .02 .02 4.34 4 _____ ± 2.054 invNorm(.98, 0, 1) = 2.054 𝑧 ∗ =2.054 _____ ± 4.457 With calculator: _________, _________ STAT TESTS ZInterval (7) Inpt: 𝜎: 𝑥 : n: C-Level: Data Stats 4.34 _____ _________, _________ 4 0.96 We are 96% confident that the interval from ______ to ______ captures the true mean final exam score, 𝜇, for the class of AP Stats students.
= 𝑧 ∗ 𝜎 𝑛 96% confidence level → 𝑧 ∗ = 2.054 Margin of Error 4=2.054 4.34 𝑛 1.95= 4.34 𝑛 1.95 𝑛 =4.34 𝑛 =2.226 𝑛=4.96 We still round up. 𝑛=5
𝜎 To know 𝜎, we would have to know all of the test scores. If we knew them, we would simplify find 𝜇 rather than estimate it. 𝑠 𝑥 𝑡-score 𝑡-distribution 𝑥 −𝜇 𝜎 𝑛 𝑥 −𝜇 𝑠 𝑥 𝑛 From now on, use 𝑧 for proportions and 𝑡 for means (if 𝜎 is unknown) 𝑧 𝑎 𝑝 𝑡 𝑎 𝑥
A 𝑡-distribution has a different shape than the Standard Normal Curve A 𝑡-distribution has a different shape than the Standard Normal Curve. It’s still symmetric with a peak at 0 but has much more area in the tails. The shape of a 𝑡-distribution depends on the sample size, which indicates its degrees of freedom: df = 𝑛−1→ 𝑡 𝑛−1 𝑡 has the same interpretation as 𝑧, “how many std dev 𝑥 is from 𝜇.”
𝒅𝒇=𝒏−𝟏 more more more closer more
SAME one-sample T interval for mean conditions SAME
wider than the 𝑧-interval calculated before. Estimate ± Margin of Error 𝑡-distribution 𝑠 𝑥 𝑛 𝑥 ± 𝑡 ∗ .96 .02 .02 4 _____ ± 3.482 Can’t use invNorm since we’re using a 𝑡-distribution. We now use Table B/𝑡-table (next slide) _____ ± _______ _________, _________ 𝑑𝑓=𝑛−1=4−1=3 wider than the 𝑧-interval calculated before. So 𝑡 ∗ =3.482 Or use invT(area on left, df) invT(0.98, 3) = 3.482 With calculator: STAT TESTS TInterval (8) Inpt: 𝑥 : 𝑆 𝑥 : n: C-Level: Data Stats _____ _____ _________, _________ 4 0.96
So 𝑡 ∗ =3.482 Notice table gives area on the right Have to look at 3 things: 1) Upper tail prob (.02) 2) 𝑑𝑓=4−1=3 3) Conf level (96%) So 𝑡 ∗ =3.482
We are 96% confident that the interval from ______ to ______ captures the true mean final exam score, 𝜇, for the class of AP Stats students.
Practice Quiz for 10 min! Trade papers and grade. 100.644 1.497 We want to estimate the true mean lap time, 𝜇, for Mr. Brinkhus’ career at MB2 Raceway at a 95% confidence level. one sample T interval for mean Random: “Think of 9 laps as a SRS.” Normal: The population distribution is normal, so the sampling distribution of 𝑥 is approximately normal. Independent: Mr. Brinkhus did more than 10 9 =90 laps.
Estimate ± Margin of Error 𝑡-distribution 𝑠 𝑥 𝑛 𝑥 ± 𝑡 ∗ .95 .025 .025 1.497 9 100.644± 2.306 𝑑𝑓=𝑛−1=9−1=8 100.644±1.151 𝑡 ∗ =invT 0.975, 8 =2.306 99.493, 101.795 We are 95% confident that the interval from 99.49 to 101.80 captures the true mean lap time, 𝜇, for Mr. Brinkhus’ career at MB2 Raceway.
112.12 86.14, 138.1 With calculator: 36.32 STAT TESTS TInterval (8) Inpt: 𝑥 : 𝑆 𝑥 : n: C-Level: Data Stats 112.12 36.32 10 0.95 Notice the interval got a lot wider because of the outlier.
Sample more laps! Make 𝑛≥30. Normal – pop dist may not be normal so sampling dist may not be normal cannot ROBUST accurate Sample more laps! Make 𝑛≥30.
𝑛≥30 CLT Central Limit Theorem outliers strong skewness You MUST graph the sample data (using a dotplot or boxplot) to see if the data is approximately normal.
Since the sample of 50 students was only from your classes and not from all students at your school, we should not use a 𝑡 interval to generalize about the mean GPA for all students at the school. Not an SRS. Since the distribution is only moderately skewed and the sample size is larger than 15, it is safe to use a 𝑡 interval. Since the sample size is small and there is a possible outlier, we should not use a 𝑡 interval.