1. Hypothesis

Question:Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

In a sample of 167 children selected randomly from one town, it is found that 37 of them suffer from asthma. At the 0.05 significance level, test the claim that the proportion of all children in the town who suffer from asthma is 11&#37;.

My problem is that I don't even know where to start at.
If someone could put me in the right direction of what to do, I would love it. Thanks

So would H(0)=11% and H(1) not equal 11%
Im not sure if it is a t or z statistic

2. This is a hypothesis test of proportions. First thing to know. Z-statistic should be used when computing test of proportions.

H0 = Population proportion = .11
H1 = Population proportion not equal to .11

This is a two-tailed test. Let p =sample proportion, and P = population proportion.

The formula for testing proportions is Z = p - P/sqrt[P*(1-P)/n] where n is the total sample size.

So P = .11 as noted in H0, and to calculate the sample proportion (which is 37/167), you have to divide 37 (asthma sufferers) by 167 (total sample) = .22

So then you plug the info. in the formula so that = .22 - .11/sqrt[(.11*.89)/167]. The numerator will be .11 (.22 - .11), and the denominator will be .02.

Divide .11 by .02 to find z-value or test-statistic = 5.5

At .05 level of significance for a two-tailed test, the table (critical) z values are between -1.96 and +1.96, any test statistic that falls outside this region, you should reject the null hypothesis. If it falls within this region, you should accept the HO. Since 5.5 is larger than 1.96, you should reject the null hypothesis. The proportion is not 11%.

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