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It is well known that a placebo, a fake medication or treatment, can sometimes have a positive effect just because patients often expect the medication or treatment to be helpful. The article "Beware the Nocebo Effect" (New York Times, Aug. 12, 2012) gave examples of a less familiar phenomenon, the tendency for patients informed of possible side effects to actually experience those side effects. The article cited a study reported in The Journal of Sexual Medicine in which a group of patients diagnosed with benign prostatic hyperplasia was randomly divided into two subgroups. One subgroup of size 55 received a compound of proven efficacy along with counseling that a potential side effect of the treatment was erectile dysfunction. The other subgroup of size 52 was given the same treatment without counseling. The percentage of the no-counseling subgroup that reported one or more sexual side effects was 15.3 %, whereas 43.6 % of the counseling subgroup reported at least one sexual side effect. State and test the appropriate hypotheses at significance level .05 to decide whether the nocebo effect is operating here. (Note: The estimated expected number of "successes" in the no-counseling sample is a bit shy of 10, but not by enough to be of great concern (some sources use a less conservative cutoff of 5 rather than 10).)

Short Answer

Expert verified

Reject null hypothesis and can conclude that there does appear to be Nocebo effect.

Step by step solution

01

To determine the large- sample test procedure

There are two subgroups. Denote with \({p_1}\)the true proportion of patients which did not receive counseling, and with \({p_2}\)the true proportion of the patients which received counseling about the possible side effect.

For the no-counseling subgroup (first subgroup) 15.3 % of 52 who received no counseling, and for the second subgroup, the subgroup of patients who received counseling 43.6% of 55 reported one or more sexual side effects.

The 15.3 % of 52 is

\(m = 52 \times \frac{{15.3}}{{100}} = 8\)

and 43.6 % of 55 is

\(n = 55 \times \frac{{43.6}}{{100}} = 24\)

This means that 8 out of 52 patients who received no counseling reported sexual side effect, and 25 out of 55 patients who received counseling reported sexual side effect.

The hypotheses of interest are \({H_0}:{p_1} - {p_2} = 0\) versus\({H_a}:{p_1} - {p_2} < 0\). Even though the samples are not large enough (and the proportions), you can use the large sample test procedure (see the hint).

A Large-Sample Test Procedure:

The two proportion z statistic - Consider test in which\({H_0}:{p_1} - {p_2} = 0\). The test statistic value for the large samples is

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \times \left( {\frac{1}{m} + \frac{1}{n}} \right)} }}\)

where

\(\hat p = \frac{m}{{m + n}} \times {\hat p_1} + \frac{n}{{m + n}} \times {\hat p_2}\)

Depending on alternative hypothesis, the P value can be determined as the corresponding area under the standard normal curve. The test should be used when\(m{\hat p_1},m{\hat q_1},n{\hat p_1}\), and \(n{\hat q_1}\) are at least 10.

The estimates of the proportions are

\(\begin{array}{l}{{\hat p}_1} = \frac{8}{{52}} = 0.153\\{{\hat p}_2} = \frac{{24}}{{55}} = 0.436\end{array}\)

Compute a pooled proportion \(\widehat p\)

\(\begin{array}{c}\hat p = \frac{m}{{m + n}} \times {{\hat p}_1} + \frac{n}{{m + n}} \times {{\hat p}_2}\\ = \frac{{52}}{{52 + 55}} \times \frac{8}{{52}} + \frac{{55}}{{52 + 55}} \times \frac{{24}}{{55}}\\ = 0.299.\end{array}\)

02

Step 2:To find the two proportion z test value

The two proportion z test value is

\(\begin{array}{c}z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \times \frac{1}{m} + \frac{1}{n}} }}\\ = \frac{{0.153 - 0.436 - 0}}{{\sqrt {(0.299 \times 0.701) \times \frac{1}{{52}} + \frac{1}{{55}}} }}\\ = - 3.2\end{array}\)

The P value is the area under the standard normal curve to the left of z because the test is lower sided (one sided). Thus

\(P = P(Z \le - 3.2) = 0.007\)

which was obtained using the table in the appendix of the book. Since

\(P = 0.007 < 0.05 = \alpha \)

Reject null hypothesis

At given significance level. You can conclude that a higher proportion of men will experience erectile dysfunction if they received consulting about the side effect of the BPH treatment, than if there were no consulting about the side effects.

Or there does appear to be nocebo effect.

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Most popular questions from this chapter

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, n individuals are selected and asked whether \((S)\)or not (F) they favor the candidate. Then after the speech the same n people are asked the same question. The responses can be entered in a table as follows:

Before

After

S

F

S

\({{\bf{X}}_{\bf{1}}}\)

\({{\bf{X}}_{\bf{2}}}\)

F

\({{\bf{X}}_{\bf{3}}}\)

\({{\bf{X}}_{\bf{4}}}\)

Where\({{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{x}}_{\bf{2}}}{\bf{ + }}{{\bf{x}}_{\bf{3}}}{\bf{ + }}{{\bf{x}}_{\bf{4}}}{\bf{ = n}}\). Let\({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}\), and \({p_4}\)denote the four cell probabilities, so that \({p_1} = P(S\) before and S after), and so on. We wish to test the hypothesis that the true proportion of supporters (S) after the speech has not increased against the alternative that it has increased.

a. State the two hypotheses of interest in terms of\({p_1},{p_2}\),\({p_3}\), and \({p_4}\).

b. Construct an estimator for the after/before difference in success probabilities

c. When n is large, it can be shown that the rv \(\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}{{\bf{X}}_{\bf{j}}}} \right){\bf{/n}}\) has approximately a normal distribution with variance given by\(\left[ {{{\bf{p}}_{\bf{i}}}{\bf{ + }}{{\bf{p}}_{\bf{j}}}{\bf{ - }}{{\left( {{{\bf{p}}_{\bf{i}}}{\bf{ - }}{{\bf{p}}_{\bf{j}}}} \right)}^{\bf{2}}}} \right]{\bf{/n}}\). Use this to construct a test statistic with approximately a standard normal distribution when \({H_0}\)is true (the result is called McNemar's test).

d. If\({{\bf{x}}_{\bf{1}}}{\bf{ = 350,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ = 150,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ = 200}}\), and\({x_4} = 300\), what do you conclude?

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\(\begin{array}{*{20}{l}}{ Energizer: }&{8.65}&{8.74}&{8.91}&{8.72}&{8.85}\\{ Ultracell: }&{8.76}&{8.81}&{8.81}&{8.70}&{8.73}\\{ Energizer: }&{8.52}&{8.62}&{8.68}&{8.86}&{}\\{ Ultracell: }&{8.76}&{8.68}&{8.64}&{8.79}&{}\end{array}\)

Normal probability plots support the assumption that the population distributions are normal. Does the data suggest that the variance of the Energizer population distribution differs from that of the Ultra cell population distribution? Test the relevant hypotheses using a significance level of .05. (Note: The two-sample \(t\)test for equality of population means gives a \(P - \)value of .763.) The Energizer batteries are much more expensive than the Ultra cell batteries. Would you pay the extra money?

Arsenic is a known carcinogen and poison. The standard laboratory procedures for measuring arsenic \(concentration (\mu g/L)\) in water are expensive. Consider the accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method (from the article "Evaluation of a New Field Measurement Method for Arsenic in Drinking Water Samples," J. of Emvir. Engr., 2008: 382-388).

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 3 19.70 1.10 0.65

2 3 10.90 0.60 0.35

Estimate for difference: 8.800

95% CI for difference: 6.498,11.102

T -Test of difference =0 (vs not = ):

T -Value =12.1 P -Value =0.001 DF=3

What conclusion do you draw about the two methods, and why? Interpret the given confidence interval. (Note: One of the article's authors indicated in private communication that they were unsure why the two methods disagreed.)

Is the response rate for questionnaires affected by including some sort of incentive to respond along with the questionnaire? In one experiment, 110 questionnaires with no incentive resulted in 75 being returned, whereas 98 questionnaires that included a chance to win a lottery yielded 66 responses ("'Charities, No; Lotteries, No; Cash, Yes," Public Opinion Quarterly, 1996: 542โ€“562). Does this data suggest that including an incentive increases the likelihood of a response? State and test the relevant hypotheses at significance level . 10 .

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). The article "A Gap Detection Tactility Test for Sensory Deficits Associated with Carpal Tunnel Syndrome" (Ergonomics, \(1995: 2588 - 2601\)) reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for\(m = 8\)normal subjects was\(1.71\;mm\), and the sample standard deviation was\(.53\); for\(n = 10\)CTS subjects, the sample mean and sample standard deviation were\(2.53\)and\(.87\), respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of\(.01\).

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