Statistics: Critical values and Standardized Tests...How 2 do it
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Statistics: Critical values and Standardized Tests...How 2 do it

[From: ] [author: ] [Date: 11-04-23] [Hit: ]
In Greene County in Pennsylvania, a random sample of 18 residents has a mean annual income of $35100 and a standard deviation of $7375. Test the claim at a = 0.01 that the mean annual incomes in Delaware and Greene counties are not the same. Assume the population variances are equal. Hint: For n > 29,......
a) Identify the claim and state H0 and Ha
b) Find the critical value(s) and identify the rejection region(s)
c) Find the standardized test statistic
d) Decide whether to reject or fail to reject the null hypothesis
e) Interpret the decision in the context of the original claim. Assume the populations have approximately normal distributions.
A random sample of 17 residents of Delaware County in Pennsylvania has a mean annual income of $35800 and a standard deviation of $7800. In Greene County in Pennsylvania, a random sample of 18 residents has a mean annual income of $35100 and a standard deviation of $7375. Test the claim at a = 0.01 that the mean annual incomes in Delaware and Greene counties are not the same. Assume the population variances are equal. Hint: For n > 29, use the last row (∞) in the t-distribution table.

-
S(D - G) = sqrt(60840000/17 + 54390625/18) = 2569.148676

H0: μD = μG
H1: μD ≠ μG, α = 0.01

Decision Rule:

If abs((35800 - 35100) - 0)/2569.148676 > tst(33; 0.995), we reject H0.

If 0.27 ≯ 2.576, we don't Reject H0.

Conclusion:

The mean annual incomes in Delaware and Greene counties are the same
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