Test of hypothesis for proportions, standard deviation
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1. A poll of 1,068 Americans reveals that 48% of the voters surveyed prefer the Democratic candidate for president. At the .05 level of significance, test the claim that at least half of all voters prefer the democratic candidate.
2. A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawn equipment manufacturer fells the estimate is too low for households in Omaha. Among 497 randomly selected homes in Omaha, 340 had one or more lawn mowers. Test the claim that the proportion with the lawn mowers is higher than 65%.
3. In one town monthly incomes for men without college degrees are found to have a standard deviation of 650$. Use a .01 significance level to test the claim that for men without college degrees in that town, incomes have a higher standard deviation when a random sample of 22 men resulted in incomes with a standard deviation of 925$
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Solution Summary
Uses z distribution and chi squared distribution for testing hypothesis on proportions and standard deviation.
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1. A poll of 1,068 Americans reveals that 48% of the voters surveyed prefer the Democratic candidate for president. At the .05 level of significance, test the claim that at least half of all voters prefer the democratic candidate.
The hypothesis that we are testing is that 50% or more voters prefer Democratic candidate
Data
Hypthesized proportion= 50%
Sample proportion= 48%
Sample size= 1068
Significance level= 0.05
1) Hypothesis
Null Hypothesis: Ho: p = 50% (:Proportion is at least 50.%)
Alternative Hypothesis: H1: p < 50% :( Proportion is less than 50.% )
Significance level=alpha (a) = 0.05 or 5%
No of tails= 1 ( Left tail )
This is a 1 tailed ( Left tail ) test because we are testing that p < 0.5
2) Decision rule
sample size=n= 1068 is large
We use z distribution
Z at the 0.05 level of significance, 1 tailed test = 1.6449
z critical = -1.6449
if sample statistic is <-1.6449 Reject Null Hypothesis, else Accept Null Hypothesis
Alternatively,
if p value is less than the significance level (= 0.05 ) Reject Null Hypothesis, else Accept Null Hypothesis
3) Calculation of sample statistics
p= 50.00%
q=1-p= 50.00%
n= 1068
sp=standard error of proportion=square root of (pq/n)= 1.5300% =square root of ( 50.% * 50.% / 1068)
sample proportion= 48.00%
z=(sample proportion-population proportion)/ standard error= -1.3072 =(48.%-50.%)/1.53%
4) Compare sample ...
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