Null hypothesis??!?! Need Help!!!!!!?

A nationwide study of American homeowners revealed that 64% have one or more lawn mowers. A lawn equipment company feels the estimate is too low for households in Omaha. A survey of 490 homes in Omaha yeilds 331 with one or more lawn mowers. At a 0.5 level of significance, test the claim that more than 64% of the homeowners in Omaha have one or more lawn mowers





State the null hypothesis


State the alternative hypothesis


Find the critical value


Find the value of the test statistic


Find the p-value

Null hypothesis??!?! Need Help!!!!!!?
I think you mean test at the 0.05 level?





Hypothesis Test for proportions:





Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p)





To test the null hypothesis of the form


H0: p = p0, or


H0: p ≥ p0, or


H0: p ≤ p0





Assuming that n*p0 %26gt; 10 and n * (1-p0) %26gt; 10 (some will say the necessary condition here is %26gt; 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then





find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n)





where pHat = X / n





The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.


H1: p ≠ p0; p-value is the area in the tails greater than |z|


H1: p %26lt; p0; p-value is the area to the left of z


H1: p %26gt; p0; p-value is the area to the right of z





If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value %26gt; α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.





The hypothesis test in this question is:





H0: p ≤ 0.64 vs. H1: p %26gt; 0.64





The test statistic is:


z = ( 0.6755102 - 0.64 ) / ( √ ( 0.64 * (1 - 0.64 ) / 490 )


z = 1.637608





The p-value = P( Z %26gt; z )


= P( Z %26gt; 1.637608 )


= 0.05075174





Since the p-value is greater than the significance level of 0.05 we fail to reject the null hypothesis and conclude p ≤ 0.64 is plausible.
Reply:Hopefully you have a graphing calculator, as this is necessary to do this problem. You should know how to do it by hand, but the calculator helps.





Ho: p = .64


Ha: p %26gt; .64 one tailed upper





one proportion z test





your critical value would be a z observed of 1.645 (not quite) which would leave 5% in the upper portion


z observed = 1.638


p value = .051 .. you fail to reject becasue it is GREATER than .05