SOLUTION: 1. Test the claim that the proportion of people who own cats is smaller than 90% at the 0.01 significance level. The null and alternative hypothesis would be:
H0:p≥0.9
H1:p<0.9
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-> SOLUTION: 1. Test the claim that the proportion of people who own cats is smaller than 90% at the 0.01 significance level. The null and alternative hypothesis would be:
H0:p≥0.9
H1:p<0.9
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Question 1175474: 1. Test the claim that the proportion of people who own cats is smaller than 90% at the 0.01 significance level. The null and alternative hypothesis would be:
H0:p≥0.9
H1:p<0.9
Based on a sample of 500 people, 83% owned cats
What is The test statistic is: (2 decimal places)
The p value (2 decimal places).
2. You wish to test the following claim ( Ha ) at a significance level of α= 0.001.
Ho:p=0.61
Ha:p>0.61
You obtain a sample of size n= 382 in which there are 240 successful observations. For this test, you should use the (cumulative) binomial distribution to obtain an exact p-value. (Do not use the normal distribution as an approximation for the binomial distribution.) The p-value for this test is (assuming Ho is true) the probability of observing... Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! Assuming randomness etc.
test stat is a z=(p hat-p)/sqrt(p*(1-p)/n)
Critical value is z< -2.33
=-.07/sqrt(.9*.1/500)
=.-.07/0.013
=-5.21
reject Ho
p-value <0.0001 (4 x 10^(-7)
Note: realistically have to use 416 or 417 for the numerator of the proportion since 83% of 500 is not an integer. It doesn't change the answer.
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Probability is 0.2490 that it is >0.61. (0.7510 that it is 239 or fewer)
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check with the normal approximation, which ought to be about this.
np=mean or 233.02
np(1-p)=variance or 90.8778
sd is sqrt(V)=9.53
z>(239.5-233.02/9.53 which is 6.48/9.53 or z > 0.68 which is prob. 0.2482
Use the answer above of 0.2490.
