Question

An advertisement for Claritin, a drug for seasonal nasal allergies, made this claim “Clear relief without drowsiness. In studies, the incidence of drowsiness was similar to placebo.” (Time, Feb. 6, 1995, p. 43). The advertisement also reported that 8% of the 1926 Claritin takers reported drowsiness as a side effect, compared with 6% of the 2545 placebo takers.

a.   Define the parameters p1 and p2 in words. Set up null and alternate hypotheses for whether the rate of drowsiness with Claritin is higher than with placebo.
b.   Calculate the number of drowsy Claritin users and the number of drowsy placebo users. Be sure to round to whole numbers of people!
c.   Calculate , p-hat pooled, the overall rate of drowsiness.
d.   Calculate SE pooled (p-hat) using the formula
e.   Calculate the z-score and p-value for the difference in proportions.
f.   Conclude whether to reject the null hypothesis or not. Make a conclusion that includes the context of the situation (your sentence should be about drowsiness rates of Claritin vs placebo).

Answer 1

(a)

Here p1 shows the true proportion of Claritin takers reported drowsiness as a side effect.

Here p2 shows the true proportion of placebo takers reported drowsiness as a side effect.

Hypotheses are:

(b)

The number of drowsy Claritin users is

The number of drowsy placebo users is

T2 n2 . P2 = 2545 . О.06 152.7 153

(c)

$Hypotheses are Level of significance α-0.05 Test is two tailed The pooled proportion is =(0+0)/(1926+2545)=0.0686 The standard error is: p (1-2)| _ + _ | = 0.0076 i ni Test Statistics: p1-p2 )-\p1-p2 ) =2.63 SE Critical value: CV = 1.64 Rejection Region: If z > 1.64, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value =0.0043 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: NORMSINV(1-0.05) -1-NORMSDIST(2.63)$

The pooled p-hat is 0.0686

(d)

The standard error is

SE = 0.0076

(e)

The z-score = 2.63

The p-value = 0.0043

(f)

Since p-value is less than 0.05 so we reject the null hypothesis. There is evidence to conclude that the rate of drowsiness with Claritin is higher than with placebo.