A.8 Answer: TW 8 tutorial

Answers for Sect. 8.2

Answers implied by H5P.

The decision-making process begins with making an assumption about the population parameter. This means we know what to expect from the sample statistic. We never know exactly what value of the statistic we will see in the sample, because of sampling variation. But we can have some of idea of what values are reasonable to expect. Then we take the sample (that is, we make the observations). Then we compare the sample statistic that we observed... to the sample statistic we expected. If what we observe is inconsistent with what was expected, then the the assumption is unlikely to be true. However, if what we observe is consistent with what was expected, then the the assumption is probably true.
Step 1: Assumption about population parameter. Step 2: Expectation for sample statistic. Step 3: Observation of sample statistic. Decision: Consistent? Conclusion A: Yes, supports assumption. Conclusion B: No, doesn't support assumption.

Answers for Sect. 8.3

$H_0$ : $p = 0.3$ 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. 8.4

The parameter of interest is the population mean pizza diameter, say $\mu$ .
From the output: $\bar{x} = 11.486$ inches and $s = 0.247$ inches.
Use $\text{s.e.}(\bar{x}) = s/\sqrt{n} = 0.247/\sqrt{125} = 0.02205479$ inches.
$s$ 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 $\text{s.e.}(\bar{x}) = 0.02205$ .
$t = (11.486-12)/0.02205 = -23.3$ .
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 $11.486\pm 0.0441$ ).
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. 8.5

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. 8.6.1

$s = 7.145\,\text{m}$ ; $\text{s.e.}(\bar{x}) = s/\sqrt{n} = 7.145/\sqrt{44} = 1.077\,\text{m}$ .
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 $18.19\,\text{m}$ .
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: $16.02\,\text{m}$ ; standard deviation: $7.145\,\text{m}$ ) of the students is not $13.1\,\text{m}$ ( $95$ % CI for population mean guess: $13.85$ to $18.19\,\text{m}$ ).
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. 8.6.2

Researchers (Nataraja et al. 1999) examined the strength of fibre reinforced concrete, by using a study design called an experiment. In batch 1, a sample of size $30$ was used; the sample mean number of blows till the first crack appeared in the test cylinders was $98$ , and the amount of variation in the number of blows was measured using the standard deviation as $54$ . Because the data are a sample, the sample mean will estimate the population mean with some sampling error.
A type of study called an experiment compared the handwriting legibility for school children (Ryan et al., 2010) having cerebral palsy when using specialist school furniture with standard school furniture (which acted as a control). They used a random sample of size $30$ from children registered at their facility in Canada. The sample mean for the difference in legibility was $-0.1$ , and a $95$ % confidence interval was from $-0.8$ to $0.6$ . Using the standard equipment, the smallest value recorded for legibility was $19$ , and the largest was $34$ , so the range was $15$ .

References

Nataraja MC, Dhang N, Gupta AP. Statistical variations in impact resistance of steel fiber-reinforced concrete subjected to drop weight test. Cement and Concrete Research. 1999;29:989–95.