Question 300909: Need help with these statistics, I am lost I need a solution A.S.A.P this is due today 5/5/10. Your help is greatly appreciated.
1) A telephone survey gives 670 consumers two choices: Do they prefer Coke or Pepsi? Exactly 158 of those surveyed state that they prefer Coke. Assuming that = 0.10, test the hypothesis that the proportion of the population that prefers Coke is 50%.
State the null and alternate hypotheses
Calculate the sample proportion
Calculate the value of the test statistic.
Determine the critical value(s).
State your decision: Should the null hypothesis be rejected?
2) An aerospace parts factory has two separate production lines making fasteners; each production line can be regarded as creating a separate population of fasteners, whose widths are normally distributed. A sample of 20 fasteners is selected from each production line for quality control inspection. The width of each selected fastener is measured; and the standard deviations s1 and s2 in the measured widths of the selected fasteners from each production line are calculated. If the design standard deviation in fastener width for each production line population is unknown, can the sample values s1 and s2 be used instead?
* Yes
* No
Answer by alanc(27)   (Show Source):
You can put this solution on YOUR website! 512
and 158,
total 670 consumers
I'll assume this is a normal distribution.
P(u = 0.5),
My significance level is the default 5%
so Find P(x = 0.5)
our u mean is 158/670 = 0.235820
our x mean estimator is 0.5
our standard deviation is s = 0.1
z = (x-u)/s
z = (0.5 - 0.23582)/0.1 = 2.6417910447761194029850746268657
P( Z <= 2.6417910447761194029850746268657) = 2 * 0.4959 = 2* 49.59 % = 99.16 %
p-value is 49.59%
The sample proportion : 158/670 = 0.23582
The critical value is z= 2.64179
Actually, the test statistic at Z = 2.64 is very far away from the mean u = 0. That means Z is in the rejection region. If I changed the statement to :
P( Z >= 2.6417910447761194029850746268657) it would equal :
P( Z > 2.64) = 1 - P(Z < 2.64) = 1 - 0.9916 = 0.0084
I must reject the null hypothesis because the real P-Value is 0.0084 which is way less than 5% significance level.
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Redoing the problem because I misunderstand what I am given. If you say the significance level is 0.1 then my work changes.
total population = n = 670
our u mean is 158/670 = 0.235820
our x mean estimator is 0.5
our standard deviation is s = sqrt(pq/n)
where p = 158/670, q = 512/670
s = sqrt(0.23582*0.76418)
s= 0.0164
still we have P(Z >= 16.1085) which is close to zero.
I still reject the null hypothesis since this p-value is way less than 0.1 significance level.
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2 separate production lines n = 20 is our sample population
measured standard deviations s1 and s2
assume mean is u1 and u2
P( Z <= (x - u1)/[s1/sqrt(20)]) = 1 - 0.05
P( Z <= (x - u2)/[s2/sqrt(20)]) = 1 - 0.05
I would say yes we can use s1 and s2 as estimators for standard deviation.
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