What should the advertising company conclude at the 5% level of significance?
What should the advertising company conclude at the 5% level of significance?
An advertising company wants to determine if the cartoons series Voltes V, appeals to male viewers more than to female viewers. Based on random telephone interviews,it was found that 23 out of 48 males and 41 out of 90 females watch the series regularly. What should the advertising company conclude at the 5% level of significance?
x(m) = 23
n(m) = 48
x(f) = 41
n(f) = 90
alfa = 0.05
The test of significance for comparison of two proportions of independent samples is performed.
When p1 = x1/n1 is greater than p2 = x2/n2, after application of the continuity correction rule the formula for z is:
[(x1- 1/2)/n1] - [(x2- 1/2)/n2]
——————- —————— = z
sqrt [p *q * ((1/n1) - (1/n2))]
Verification whether it is appropriate to employ the test based on the normal distribution:
The rate of those who reported that watch the series regularly:
p = (23+ 41) / (48 + 90) = 64 /138 =0.4638
The rate of those who do not watch regularly:
q = 1 - p = 1 - 0.4638 = 0.5362
Since n(m) = 48 and n(f) = 90, all these four values are above 5:
n(m) * p
n(m) * q
n(f) * p
n(f) * q
Thus it is possible to use the test based on the normal distribution.
Since the proportion of persons who watch the series regularly is greater in the group of males (0.479 in comparison with 0.456 for females), the numerator in the formula for z will be:
numerator = [(23 - 0.5)/48] - [(41 +0.5) / 90] = 0.46875 - 0.46111 = 0.00759
denominator = sqrt [0.4638 * 0.5362 * ((1/48) + (1/90))] = sqrt 0.00794 = 0.08913
z = 0.00759 / 0.08913 = 0.0852
The two-tailed normal tables give a probability of 0.9321 or 93.21 %.
This P > 0.05 (or 5 %).
The conclusion is that the rate of male viewers does not differ significantly from that of females; in other words, the observed difference in rates can be explained by chance or sampling variation.
November 28th, 2009 at 10:41 am
x(m) = 23
n(m) = 48
x(f) = 41
n(f) = 90
alfa = 0.05
The test of significance for comparison of two proportions of independent samples is performed.
When p1 = x1/n1 is greater than p2 = x2/n2, after application of the continuity correction rule the formula for z is:
[(x1- 1/2)/n1] - [(x2- 1/2)/n2]
——————- —————— = z
sqrt [p *q * ((1/n1) - (1/n2))]
Verification whether it is appropriate to employ the test based on the normal distribution:
The rate of those who reported that watch the series regularly:
p = (23+ 41) / (48 + 90) = 64 /138 =0.4638
The rate of those who do not watch regularly:
q = 1 - p = 1 - 0.4638 = 0.5362
Since n(m) = 48 and n(f) = 90, all these four values are above 5:
n(m) * p
n(m) * q
n(f) * p
n(f) * q
Thus it is possible to use the test based on the normal distribution.
Since the proportion of persons who watch the series regularly is greater in the group of males (0.479 in comparison with 0.456 for females), the numerator in the formula for z will be:
numerator = [(23 - 0.5)/48] - [(41 +0.5) / 90] = 0.46875 - 0.46111 = 0.00759
denominator = sqrt [0.4638 * 0.5362 * ((1/48) + (1/90))] = sqrt 0.00794 = 0.08913
z = 0.00759 / 0.08913 = 0.0852
The two-tailed normal tables give a probability of 0.9321 or 93.21 %.
This P > 0.05 (or 5 %).
The conclusion is that the rate of male viewers does not differ significantly from that of females; in other words, the observed difference in rates can be explained by chance or sampling variation.
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