The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We interested in testing to determine whether the variance of the population is significantly more than 0.003. 1. Write Null and Alternative hypotheses 2. Compute Test Statistics 3. Rejection (Decision) Rules (using alpha = 0.05) a) P-value approach b) Critical Value approach 4. Conduct 95% confidence interval

(1) Null hypothesis: The variance of the population is 0.003.

Alternate hypothesis: The variance of the population is greater than 0.003.

(2) The test statistic is 0.051

(3)a P-value approach: Reject the null hypothesis if the P-value is less than or equal to the significance level 0.05.

(b) Critical value approach: Reject the null hypothesis if the test statistic is less or greater than the critical value 1.645.

(4) 95% confidence interval for the population variance is between a lower limit of -0.019 and an upper limit of 0.025.

Step-by-step explanation:

(1) A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from a population parameter which is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

(2) Test statistic (z) = (sample variance - population variance) ÷ sample sd/√n

sample variance = (sample sd)^2 = 0.06^2 = 0.0036

population variance = 0.003

sample sd = 0.06

n = 26

z = (0.0036 - 0.003) ÷ 0.06/√26 = 0.0006 ÷ 0.0118 = 0.051

(3) a Using the p-value approach, the null hypothesis is rejected if the p-value is less than or equal to the significance level.

(b) Using the critical value approach, the null hypothesis is rejected if the test statistic is less or greater than the critical value which is 1.645 at 0.05 significance level.

(4) Confidence interval = population variance + or - Error margin (E)

population variance = 0.003

population sd = √0.003 = 0.055

n = 26

degree of freedom = n - 1 = 26 - 1 = 25

t-value corresponding to 25 degrees of freedom and 95% confidence interval = 2.060

E = t×sd/√n = 2.060×0.055/√26 = 0.022

Lower limit = population variance - E = 0.003 - 0.022 = -0.019

Upper limit = population variance + E = 0.003 + 0.022 = 0.025

95% confidence interval for the population variance is between -0.019 and 0.025.

x = 25/2304

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11