3. Test the null hypothesis that the variance is equal to 40 against the alterna
ID: 3321446 • Letter: 3
Question
3. Test the null hypothesis that the variance is equal to 40 against the alternative that it is not. a. Define the null and alternative hypothesis. b. Define the rejection region and state the rejection rule. c. Compute the test statistic. d. Conclusion. e. How would the test statistic change if we collected a bigger sample? f. How would the rejection region change? 4. Compute the number of ways you can select 2 elements from 8 elements. 5. Suppose x is a binomial random variable with n=5 and p=.4 what is the probability that x is equal to 2. 6. Assume that x has an exponential distribution with equal to 4 What is the mean of x? What is the standard deviation of x?_What is the probability that x is greater than 2?Explanation / Answer
Question 3
Here, we have to use Chi square test for population variance.
Part a
The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The population variance is 40.
Alternative hypothesis: Ha: The population variance is not 40.
H0: 2 = 40 versus Ha: 2 40
Part b
Rejection region and rejection rule is given as below:
We assume 5% level of significance. = 0.05, /2 = 0.05/2 = 0.025
We are given
Sample standard deviation = S = 6
Sample size = n = 25
Degrees of freedom = df = n – 1 = 25 – 1 = 24
Lower critical value = 12.4012
Upper critical value = 39.3641
Rejection region = values < 12.4012 and values > 39.3641 (outside the interval (12.4012, 39.3641))
Rejection rule: Reject the null hypothesis if Chi square test statistic is less than 12.4012 or greater than 39.3641.
Part c
Test statistic formula is given as below:
Chi square = (n – 1)*S2 / 2
We are given
Sample standard deviation = S = 6
Sample size = n = 25
Degrees of freedom = df = n – 1 = 25 – 1 = 24
Chi square = (25 – 1)*6^2/40
Chi square = 21.6000
Part d
We have
Chi square = 21.6000
Degrees of freedom = df = n – 1 = 25 – 1 = 24
= 0.05
So,
P-value = 0.3969
P-value > = 0.05
So, we do not reject the null hypothesis that the population variance is 40.
There is sufficient evidence to conclude that the population variance is 40.
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