Question

Instruction for Hypothesis Testing (Problems 17-25, Except 24) For One Population: Show your work for finding Test value, Critical value(s), or P-value. For Two Populations: Use your calculator to give the values of Test Value and P-value. O O O O Indicate your parameter Indicate the claim List the given information Then start your five steps. Pay attention to the method, such as Traditional or P- value Put the step numbers inside a box For step (1): Indicate Ho Indicate Hi Tell if Ho or Hy is the claim (put the word claim inside parentheses in front of the claim sentence) o Identify the type of test such as right-tail test, left-tail test, or two-tail test The name of the test is in the direction of H1 For step (2): Find Critical Value (CV) for Traditional method Find Test value for P-value method For step (3): Find Test value for Traditional method Find P-value for P-value method For step (4), decide to reject Ho or not reject Ho. Circle your result. For Traditional method, draw the bell curve. Separate critical region from non-critical region using CV and locate your test value on the graph For P-value method, compare your P-value with a. For step (5), write your summary according to the slide #5 of lesson 8.2. Other similar statements are not expectable! On this step: Underline is or is not, also underline reject or support. . O O . O O . o O o

Answer 1

20. Claim : The variance of the sugar content of its yogurt is less than or equal to 25 that is $\sigma^2 \leq 25$

The claim contains less than or equal to so it comes under the null hypothesis and the alternative hypothesis contains exact opposite sign of null

The opposite sign of less than or equal to is greater than.

Parameter: Population variance $(\sigma^2)$

Claim: $\sigma^2 \leq 25$

Given information:

n = sample size = 20

s² = sample variance = 36

$\sigma^2=25$

alpha = 0.10

Step (1)

$H_0: \sigma^2\leq 25 \ (claim)$

$H_1: \sigma^2 > 25$

The alternative hypothesis contains greater than sign, so the test is a right-tailed test.

To test the population variance chi-square test is used.

Step (2): Critical value

Alpha = 0.10

Degrees of freedom = n - 1 = 20 - 1 = 19

The test is right tailed test.

The critical value using the chi-square table with alpha 0.10 in right tailed area and DF 19 is 27.204

Step (3): Test statistics

The formula of chi-square test statistics is,

$\chi ^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(20-1)36}{25}=27.36$

Test statistics = 27.36

Step (4)

Critical region 0.10 X=27.204

Decision rule: If the test statistics falls in critical region then reject the null hypothesis otherwise fail to reject the null hypothesis.

Test statistics 27.36 falls in critical region that is at right side of the criitcal value 27.204 so reject the null hypothesis.

Reject the null hypothesis

Step (5)

Conclusion: Reject the null hypothesis, that means there is no sufficient evidence to support the claim that the variance of the sugar content of it's yogurt is less than or equal to 25.