Question

A study was conducted on 50 pairs on twins. In each pair:

The stress level for each study participant was recorded as a score from 0 to 100.

pair twin1 twin2 difference
1 75.25909 65.98412 9.274965
2 43.47533 46.69892 -3.223586
3 76.59599 100.00000 -23.404015
4 75.44859 34.13983 41.308760
5 58.29283 56.35368 1.939153
6 19.20100 65.50424 -46.303241
7 31.42866 76.54730 -45.118640
8 44.10559 52.82129 -8.715704
9 49.88466 0.00000 49.884656
10 98.09307 20.09157 78.001499
11 65.27187 68.76187 -3.489998
12 34.01982 57.66864 -23.648821
13 27.04686 29.78053 -2.733665
14 44.21077 54.52524 -10.314472
15 44.01570 33.55094 10.464759
16 41.76978 65.26790 -23.498121
17 55.04447 15.24705 39.797421
18 32.16158 68.97823 -36.816656
19 58.71367 65.45238 -6.738714
20 25.24923 59.95865 -34.709414
21 45.51464 58.57469 -13.060049
22 57.54791 65.72015 -8.172238
23 52.66673 38.52970 14.137030
24 66.08379 54.42494 11.658853
25 48.85786 77.92407 -29.066206
26 60.07216 91.02907 -30.956914
27 71.71539 62.31314 9.402243
28 36.18092 54.46739 -18.286469
29 24.30801 30.63795 -6.329936
30 50.93452 14.87241 36.062111
31 45.28587 34.08731 11.198555
32 39.14223 95.62249 -56.480258
33 41.33379 81.16427 -39.830472
34 37.01057 51.41453 -14.403964
35 64.53501 45.25569 19.279323
36 73.03824 45.43060 27.607638
37 69.84321 87.90961 -18.066398
38 41.40974 49.72666 -8.316921
39 74.76608 95.68056 -20.914482
40 44.41307 77.40023 -32.987157
41 85.15806 96.97937 -11.821307
42 61.21492 31.80214 29.412785
43 40.94432 58.25113 -17.306808
44 33.35913 31.57385 1.785286
45 26.66859 75.88777 -49.219181
46 28.68819 61.59153 -32.903341
47 18.72436 49.53478 -30.810425
48 73.13074 100.00000 -26.869260
49 66.64094 64.87059 1.770355
50 45.45343 87.89632 -42.442892

The difference variable in the data set represents stress level of twin 1 minus stress levels of twin 2. Here are the sample mean and standard deviation for the difference.

mean(difference)
## [1] -7.679487
sd(difference)
## [1] 27.98065
  1. You want to test if sport influences average stress level. State null and alternative hypotheses in terms of the population average of the difference \(\mu_d\).

\(H_0:\) \(\mu_d = 0\), i.e. twin 1 and twin 2 have the same stress levels

\(H_a:\) \(\mu_d \neq 0\), i.e. twin 1 and twin 2 have different stress levels

  1. Compute the 90% confidence interval for the difference.

\([\bar{d}-1.68\frac{s_d}{\sqrt{n}}, \bar{d}+1.68\frac{s_d}{\sqrt{n}}] = [-14.31, -1.05]\)

  1. Does the confidence interval cover zero? What conclusion can be drawn?

It does not. Thus we can reject \(H_0\) in favor of \(H_a\) and we are 90% confident that sport impacts stress level.

  1. Now find the test statistics and estimate the p-value.

\(t_{obs}=\frac{\bar{d}-0}{s_d/\sqrt{n}} = -1.94\)

p-value = \(P(|T|>|t_{obs}|) = 0.06\)

  1. Draw a conclusion using p-vlaue from 4. What significance level we need to use to make the results consisted with the conclusion from 3?

Significance level \(\alpha = 0.1\). As p-value \(< 0.1\) we can reject \(H_0\) in favor of \(H_a\).

  1. Perform statistical testing to check if sport reduces the stress level, use significance level \(\alpha = 0.01\).

\(H_0:\) \(\mu_d = 0\), i.e. twin 1 and twin 2 have the same stress levels

\(H_a:\) \(\mu_d < 0\), i.e. twin 1 has lower stress level than twin 2

p-value = \(P(T<t_{obs}) = 0.03\)

p-value > 0.01, we cannot reject \(H_0\) in favor of \(H_a\). We do not have enough evidence to say that sport reduces the stress level with 99% confidence.

  1. Suppose we know that sport has no influence on stress. What type of error did we make in 3?

type I error = reject null if null is true

  1. Provide the reason why we may made this error.

Significance level \(\alpha = 0.1\) implies that type I error will appear in 10% of the studies.

  1. Suppose we know that sport reduces the stress level. What type of error did we make in 6?

type II error = fail to reject null if alternative is true

  1. How can we reduce the probability to make this error?

Increasing \(\alpha\) will increase the power of statistical test and decrease the chance to make type II error.