Question

Please answer questions 4 and 6 with steps.

4. Hypothesis Test For the sample data from Exercise 1, we have s = 0.5 and we want to use that sample to test the claim that the sample is from a population with a standard devia- tion less than 1.8 min (as it was in the past with multiple waiting lines). a. Identify the null and alternative hypotheses. b. Find the value of the test statistic. c. Using technology, the P-value is found to be 0.0001. What should we conclude about the null hypothesis? d. Given that the P-value is 0.0001, what should we conclude about the original claim? e. What does the conclusion suggest about the effectiveness of the change from multiple wait- ing lines to a single line? Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population. 5. Cans of Coke Data Set 19 in Appendix B includes volumes (oz) of a simple random sample of 36 cans of regular Coke. Those volumes have a mean of 12.19 oz and a standard deviation of 0.11 oz, and they appear to be from a normally distributed population. If we want the filling process to work so that almost all cans have volumes between 11.8 oz and 12.4 the range rule of thumb can be used to estimate that the standard deviation should be less than 0.15 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.15 oz. Use a 0.05 significance level. 6. Cans of Pepsi Repeat the preceding exercise using these statistics from a simple random sample of cans of regular Pepsi: n = 36, x = 12.29 oz, s = 0.09 oz. OZ,

Sample data from Exercise 1 used for question 4 :

6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7

data set for question 6

Answer 1

Solution :

4(a) The null and alternative hypotheses would be as follows :

$\large H_{0}: \sigma^{2}\geq (1.8)^{2}$

$\large H_{1}: \sigma^{2} < (1.8)^{2}$

4(b) To test the hypothesis the most appropriate test is chi-square test for single variance. The test statistic is given as follows :

$\large \chi^{2} = \frac{(n-1)s^{2}}{\sigma^{2}}$

Where, s is sample standard deviation, n is sample size and σ is hypothesized value of population standard deviation under H_{​​​​​​0}.

We have, s = 0.5, n = 10 and  σ = 1.8

x? (10 – 1) (0.5) (1.8)2 = 0.6944

The value of the test statistic is 0.6944.

c) The p-value for the test is 0.0001.

Significance level = 0.05

(0.0001 < 0.05)

Since, p-value is less than the significance level of 0.05, therefore we shall reject the null hypothesis at 0.05 significance level.

d) At 0.05 significance level, there is sufficient evidence to support the claim that sample is from a population with a standard deviation less than 1.8.

e) The conclusion suggest that the single waiting line is effective in reducing the variability of waiting times.