One- and Two-Sample Tests of Hypothesis, Variance, and Chi-squared Analysis Trouble Sets

•Exercises 19 and 20 (Ch. 17)

Part 10

31. A new weight-watching company, Excess weight Reducers International, advertises those who sign up for will lose, within the average, 15 pounds the first 2 weeks with a regular deviation of two. 8 pounds. A arbitrary sample of 50 people who signed up with the new weight-loss program uncovered the mean loss to be 9 pounds. At the. 05 level of value, can we determine that those joining Weight Reducers on average will lose less than twelve pounds? Identify the p-value. Answer

H0: Mean weight lose = twelve pound (µ = 10)

H1: Mean loosing weight < twelve pound (µ < 10)

Test Statistic used is Z test out. Given that = 9, n= 50, = 2 . 8 Decision rule: Decline the null hypothesis, in the event the p worth is less than the significance level 0. 05. Specifics

Z Test of Speculation for the Mean

Data

Null Hypothesis =10

Level of Significance0. 05

Population Standard Deviation2. 8

Sample Size50

Test Mean9

Intermediate Calculations

Standard Error from the Mean0. 395979797

Z Check Statistic-2. 525381361

Lower-Tail Check

Reduce Critical Value-1. 644853627

p-Value0. 00577864

Decline the null hypothesis

P benefit = P (Z < -2. 5253) = 0. 00578

Conclusion: Reject the null hypothesis, since the P-value is less than the importance level. The sample delivers enough data to conclude that weights shed is less than 15 pounds. 32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced blueberry is being overfilled. Assume the normal deviation in the process is definitely. 03 ounces. The quality control department had taken a randomly sample of 50 cans and located that the arithmetic mean excess weight was of sixteen. 05 ounces. At the 5% level of value, can we conclude that the mean weight is definitely greater than sixteen ounces? Identify the p-value. Answer

The null speculation tested is usually

H0: Suggest weight from the can of sliced pineapple = 18 ounces. (µ = 16) The alternative hypothesis is

H1: Imply weight in the can of sliced blueberry > 18 ounces. (µ > 16) Significance level =0. 05

The test figure used is definitely. Given that sama dengan 16. 05, n= 50, = 0. 03 Rejection standards: Reject the null hypothesis, if the calculated value of the test statistic is higher than the critical value with the 0. 05 significance level. Details

Z . Test of Hypothesis to get the Imply

Data

Null Hypothesis =16

Level of Significance0. 05

Inhabitants Standard Deviation0. 03

Sample Size50

Sample Mean16. 05

Intermediate Calculations

Standard Problem of the Mean0. 004242641

Unces Test Statistic11. 78511302

Upper-Tail Test

Upper Crucial Value1. 644853627

p-Value0

Deny the null hypothesis

Conclusion: Decline the null hypothesis, since the calculated worth of check statistic can be greater than the critical benefit. The test provides enough evidence to back up the claim which the mean pounds of the can easily is more than 16 oz .. P-value sama dengan P (Z > 10. 78511302) sama dengan 0.

32. A recent content in the Wall Street Journal reported that the 30-year type of loan is now below 6 percent. A sample of eight tiny banks inside the Midwest uncovered the following 30- year costs (in percent):

4. 8 five. 3 6th. 5 4. 8 6. 1 a few. 8 6th. 2 five. 6

At the. 01 significance level, can we conclude that the 30-year mortgage rate to get small financial institutions is less than six percent? Estimation the p-value. Answer

The null hypothesis tested can be

H0: The 30-year mortgage rate for small banking companies is = 6 % (µ = 6%) The alternative hypothesis is usually

H1: The 30-year type of mortgage for small banks is < 6th % (µ < 6%) Significance level = 0. 01

Test out Statistic applied is.

Decision secret: Reject the null speculation, if the benefit of test out statistic is no more than the crucial value of t in 7 df. Details

t Check for Hypothesis of the Indicate

Data

Null Hypothesis =6

Level of Significance0. 01

Test Size8

Sample Mean5. 6375

Sample Normal Deviation...