to use a chi-square test for testing variance (o2) and standard deviation (a: .

Question

to use a chi-square test for testing variance (o2) and standard deviation (a: . Verify The sample is random. The population is normally distributed. 2. a. State the claim both verbally and mathematically b. Identify the null and alternative hypotheses. (State Ho and Ha) 3. Determine what type of test you will be doing by the sign in the alternative hypothesis: If Ha is then right tailed If Ha is+ then two tailed 4. Specifys Lhe sample size, n The sample variance, s . The population variance, ?2 . The level of significance, a. . The degrees of freedom, df:--I 5. Determine the critical value(s) by using the chi-square distribution (table 6) on your formula packet. For a left tailed test, look up the d.f with 1-a. This critical value is x. For a right tailed test, look updf. with ?. This critical value is d. For a two tailed test, look up d. f. with I-½ a for the left tail and ½ a for the right tail. This will give you two critical values oft? and . . 6. Determine the rejection region(s) based on the critical valuc(s). . . . For left tailed test, the rejection region is everything less than For a right tailed test, the rejection region is everything greater than For a two tailed test, there are two rejection regions: everything lessthan ? than and everything greater .? 7. Find the standardized test statistic: ? 8. Make a decision to reject or fail to reject the null hypothesis. Reject Ho if x2 is in the rejection region. Fail to reject Ho if x2 is not in the rejection region. 9. Interpret the decision in the context of the original claim.

Explanation / Answer

As we are not supposed to answer assignement question let me take Q26. Here' the detailed answer, please reach out to me if haven't understood any part of it.

Q26)

Claim: stdev for grade 12 students on a vocab assessment test is greater than 45

n = 25

So, lets calculate test statistic = X^2 = (n-1)*S^2/Sigma^2 = (25-1)*(46/45) = 24.533

Lets now calculate critical value at alpha = .05,

For Chi-Square Formula. Degrees of freedom (df) = n-1 where n is the number of classes.

So, df = n-1 = 25-1 = 24

At df of 24 we have Critical value as 42.98

Since critical value is more than our Test statistic of 24.533 we conclude that

" There Isn't enough evidence to support the CLAIM"

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