A.9 Answer: TW 9 tutorial

Answers for Sect. 9.2

The parameter of interest is the population mean pizza diameter, say \(\mu\).
From the output: \(\bar{x} = 11.486\) and \(s = 0.247\).
Use \(\text{s.e.}(\bar{x}) = s/\sqrt{n} = 0.247/\sqrt{125} = 0.02205479\).
\(s\) tells us the variation in diameter from pizza to pizza. \(\text{s.e.}(\bar{x})\) tells us how much the sample mean is likely to vary from sample to sample, in sample of size \(125\).
The hypotheses are:

\(H_0\): The mean diameter is \(12\) inches; or \(\mu = 12\).
\(H_1\): The mean diameter is not \(12\) inches; or \(\mu \ne 12\)
Two-tailed, because the RQ asks if the diameter is \(12\) inches, or not.
The normal distribution has a mean of \(12\), and a standard deviation of \(\text{s.e.}(\bar{x}) = 0.02205\).
\(t = -23.3\).
\(t = (11.486-12)/0.02205 = -23.3\).
The \(t\)-value is huge (and negative), so the \(P\)-value is very small. (Two-tailed \(P<0.001\) from the table or from software.)
Very strong evidence exists in the sample (\(t = -23.3\); \(\text{df} = 124\); two-tailed \(P\) less than \(0.001\)) that the population mean pizza diameter of pizzas from Eagle Boys' is less than \(12\) inches (sample mean diameter: \(11.49\) inches; std. dev.: \(0.246\); \(95\)% CI from \(11.44\) to \(11.53\) inches; approximate 95% CI is \(11.486\pm 0.0441\)). (Note: The CI was computed in the Week 8 tutorial; or use \(t = 1.98\).)
Since \(n\) is much larger than \(25\), we do not require that the population has a normal distribution, just that the population is not grossly non-normal; the sample means will still have an approximate normal distribution.
Very unlikely!

Answers for Sect. 9.3

Use the \(68\)--\(95\)--\(99.7\) rule to approximate the \(P\)-values:

Less than \(0.05\); Greater than \(0.05\); Less than \(0.003\); Greater than \(0.05\); Greater than \(0.05\); Very small.

Answers for Sect. 9.4

Before and after measurement on same amputee.
\(H_0\): \(\mu_d = 0\), differences defined as 'with' minus 'without'. \(H_1\): \(\mu_d > 0\) the way I defined differences.

We are explicitly seeking to test for an improvement (i.e., walk longer distances), so the test is one-tailed. (Differences in the other direction are also OK; the signs of the differences and some other subsequent things change.)
The standard error of the mean is the standard deviation of the sample mean increase in walking distance, a measurement of how precisely the sample mean difference estimates the population mean difference.
Mean increase: \(\bar{x} = 24.9\) m; Standard deviation: \(s = 13.034\) m; standard error: \(4.122\) m.
\(t = 6.041\) and since \(t\) is very large, expect \(P\) to be very small. (The one-tailed \(P\) less than \(0.0005\) from tables.) The differences have just been defined in the opposite directions. (Notice the \(P\)-values are the same.)
Very strong evidence exists in the sample (paired \(t = 6.041\); one-tailed \(P\) less than \(0.0005\)) that the population mean 2MWT are higher after receiving the implant compared to without the implant (mean difference: \(24.9\)m higher after receiving the implant; standard deviation: \(13.034\) m; \(95\)% CI from \(15.58\) to \(34.224\)m).
The population of differences has a normal distribution, and/or \(n > 25\) or so.
Since \(n < 25\), must assume the population of differences has a normal distribution. The histogram suggests this is not unreasonable.

Answers for Sect. 9.6

\(H_0\): \(p = 0.3\) and \(H_1\): \(p > 0.3\). Here, \(n = 35\) and \(\hat{p} = 12/35 = 0.3428571\). Then,
\[ \text{s.e.}(\hat{p}) = \sqrt{\frac{0.3 \times 0.7}{35}} = 0.07745967, \] (make sure to use \(p\) not \(\hat{p}\) in the standard error calculation for the test!), and so
\[ z = \frac{0.3428571 - 0.3}{0.07745967} = 0.553. \] This is very small---less than one standard deviation from the mean---so the \(P\)-value will be quite large:

The data present no evidence (\(z = 0.55\); \(P\) large) that computer programmers are more likely to wear contact lenses (\(\hat{p} = 0.342\); approx. \(95\)% CI from \(0.182\) to \(0.503\)).

For statistical validity, we require that the number of people with and without contact lenses to be at least \(5\), which is true (\(12\) and \(23\) respectively). For external validity, the sample must be a random sample of Swedish programmers.

(For the CI, remember to use \(\hat{p}\) in the standard error calculation: \(\text{s.e.}(\hat{p}) = 0.0802329\).)

Answers for Sect. 9.7.1

\(s = 7.145\) m; \(\text{s.e.}(\bar{x}) = s/\sqrt{n} = 7.145/\sqrt{44} = 1.077\) m.
The first measures variation in the original data; the second measures precision of the sample mean when estimating the population mean.
The CI is from \(13.85\) to \(18.19\) m.
Strong to moderate evidence exists in the sample (one sample \(t = 2.714\); \(\text{df} = 43\); two-tailed \(P = 0.010\)) that the population mean guess (mean: \(16.02\) m; standard deviation: \(7.145\) m) of the students is not \(13.1\) m (\(95\)% CI for population mean guess: \(13.85\) to \(18.19\) m).
The population of differences has a normal distribution, and/or \(n > 25\) or so.
Since \(n > 25\), all OK if the histogram isn't severely skewed. Probably OK.
Not really! But was this just due to using metric units? Perhaps students are just very poor at estimating widths whatever the units being used... In fact, the Professor also had the students estimate the width of the hall in imperial units also, as a comparison.

Answers for Sect. 9.7.2

See Table A.4.

TABLE A.4: Which terms apply when? Tick the appropriate places in the table
	Applies to \(\mu\)	Applies to \(\bar{x}\)
Has a standard error	No	YES
Refers to the population	YES	No
Refers to the sample	No	YES
Value is known before the sample is taken	YES	No
Value is unknown until sample is taken	No	YES
Value is estimated	YES	No

Answers for Sect. 9.7.3

The difference in odds is statistically significant, so either 2. or 4. is consistent with this conclusion... and we cannot tell which. The size of the OR is not the issue here: if the sample OR is statistically different from \(1\), it does not tell us much about the population OR except that it is (most likely) not one.

The odds are quite different, but not statistically significant, so chance cannot be ruled out as the reason why the OR is different from one. (This likely has a smaller sample size.)
The odds are a little different, but not statistically significant, so chance cannot be ruled out as the reason why the OR is different from one.

Answers for Sect. 9.7.4

Before and after measurement on same runner.
\(H_0\): \(\mu_d = 0\), differences defined as 'after' minus 'before'. \(H_1\): \(\mu_d > 0\) the way I defined the differences.
We are explicitly seeking to test for an increase, so the test is one-tailed. (Diffs in the other direction are also OK; the signs of the differences and some other subsequent things change.)
Mean increase: \(\bar{x} = 18.73\) pmol/litre; Standard deviation: \(s = 8.3297\) pmol/litre; standard error: \(2.511\) pmol/litre.
\(t = 7.46\) and \(\text{df} = 11 - 1 = 10\). Since \(t\) is very large, expect \(P\) to be very small. (The one-tailed \(P\) less than \(0.0005\) from tables.)
Very strong evidence exists in the sample (paired \(t = 7.46\); \(\text{df} = 10\); one-tailed \(P\) less than \(0.0005\)) that the population mean \(\beta\) plasma concentrations are higher after the race compared to before the race (mean difference: \(18.73\) pmol/litre higher after the race; std. dev.: \(8.33\) pmol/litre; \(95\)% CI from \(13.14\) to \(24.33\) pmol/litre).
The population of differences has a normal distribution, and/or \(n > 25\) or so.
Since \(n < 25\), must assume the population of differences has a normal distribution. The histogram suggests this is not unreasonable.