Statement: About 30% of people cannot detect any odor when they sniff the steroi

Question

Statement:

About 30% of people cannot detect any odor when they sniff the steroid androsterone, but they can become sensitive to its smell if exposed to the chemical repeatedly. Does this change in sensitivity happen in the nose or in the brain? Mainland et al. (2002) exposed one nostril of each of 12 non-detector participants to androsterone for short periods every day for 21 days. The other nostril was plugged and had humidified air flow to prevent androsterone from entering. After the 21 days, the researchers found that 10 of 12 participants had improved androsterone-detection accuracy in the plugged nostril, whereas two had reduced accuracy. This suggested that increases in sensitivity to androsterone happen in the brain rather than the nostril, since the epithelia of the nostrils are not connected. The authors conducted a statistical hypothesis test of whether accuracy in fact did change. Let p refer to the proportion of non-detectors in the population whose accuracy scores improve after 21 days. Under the null hypothesis, p=0.5 (as many participants should improve as deteriorate in their accuracy after 21 days.)

Questions:

(A) Did the authors carry out a one- or two- sided test? What justification might they provide?

(B) The accompanying figure [presented on page 174 of the text] shows the null distribution for the number of participants out of 12 having an improved accuracy score. The probability of each outcome is given above bars. To what do these probabilities refer?

(C) What is the test statistic for the test?

(D) What is the p-value for the test?

(E) What is the appropriate conclusion? Use significance level = 0.05.

Explanation / Answer

SOL)

x = 10 , n = 12 , phat=x/n = 0.833

p0 = 0.5

a.
Since author is conducting a statistical hypothesis test of whether accuracy is changed hence it is a two-sided test.

H0: p = 0.5
H1: p is not equal to 0.5

b.
standard error, SE = sqrt(p0(1-p0)/n)
SE = sqrt(0.5(1-0.5) / 12) = 0.14434

test statistic, z = (phat-p0) / SE
z = (0.833-0.5) / 0.14434 = 2.31

c.
Since it is a 2-tailed test, hence
P_value = 2*P(Z>|z|) = 2 * P(Z> 2.31 )
= 2*(1 - P(Z< 2.31 ))
= 2*(1 - 0.9896 )
= 0.0208

d.
alpha,a = 0.05
Since P_value < alpha, hence reject H0 and conclude that accuracy is changed.

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