# Part M Course Task Math 533

Course Project Part B

a.

the average (mean) annual income was less than 50 dollars, 000

Null and Substitute Hypothesis

H0: mu= 50 (in thousands)

Ha: mu< 50 (in thousands)

Amount of Significance

Standard of Significance =. 05

Test Statistic, Critical Value, and Decision Rule

As alpha =. 05, z< -1. 645, which is decrease tailed

Being rejected region is usually, z< -1. 645

Determine test statistic, x-bar=43. 74 and s=14. 64

Z=(43. 74-50)/2. 070=-3. 0242. 070 is worked out by: s/sq-root of in

Decision Rule: The calculated test figure of -3. 024 really does fall in the rejection location of z< -1. 645, therefore I could reject the null and say there exists sufficient data to indicate mu< 50.

Interpretation of Results and Summary

p-value=. 001

. 001<. 05

Because the p-value of. 001 is less than the value level of. 05, I will decline the null hypothesis at 5% level.

95% CI=(39. 68, 47. 80)- I am 95% confident that the true suggest income is between $39, 680 and $47, 800.

Minitab Output:

One-Sample Z

Test of mu sama dengan 50 as opposed to < 55

The presumed standard deviation = 18. 64

95% Top

N Imply SE Mean Bound Z P

50 43. 74 2 . '07 47. 12-15 -3. 02 0. 001

b. the real population amount of customers who have live in a great urban region exceeds forty percent

22 people of the 50 selected live in an Urban community, which is 44%. My point estimate is definitely. 44.

Null and Alternate Hypothesis

H0: p=. forty five

Ha: p>. 40

Level of Significance

Amount of Significance=. 05

Test Figure, Critical Benefit, and Decision Rule

Since alpha=. 05, z> 1 ) 645, which is upper tailed

Rejection location is, z> 1 . 645

To carry out a large sample z-test, I need to first determine if the test size is adequate. nPo= 50(. 40)= 20 and 50(1-. 40)=30

Both are larger than 12-15, so we conclude the sample size is large enough to conduct the large sample z test.

Z=(. 44-. 40)/. 06928=. 5774. 06928 is calculated simply by sq-root ((. 4)(. 6))/50)=. 06928

Decision Rule: The calculated test statistic of. 5774 would not fall in the rejection area of z> 1 . 645, therefore I may not reject H0. There is not enough evidence in conclusion the true populace of customers whom live in urban communities is greater than 40%.

Model of Outcomes and Results

p-value=. 282

. 282>. 05

Because the p-value of. 282 is greater than the significance amount of. 05, Let me not deny the null.

95% CI= (. 299907,. 587456)- I am 95% confident the true populace proportion of customers who live in an downtown area is between 30% and 59%.

Minitab Outcome

Test and CI for One Amount

Test out of s = zero. 4 compared to p > 0. 4

95% Lower

Test X D Sample l Bound Z-Value P-Value

you 22 55 0. 440000 0. 324532 0. 58 0. 282

Using the usual approximation.

c. the average (mean) number of years lived in the current home is less than 13 years

Null and Alternate Hypothesis

H0: mu=13

': mu< 13

Level of Value

Level of Significance=. 05

Check Statistic, Crucial Value, and Decision Rule

Since alpha=. 05, z< -1. 645, which is reduce tailed.

Rejection region can be z< -1. 645

Compute the test figure, x-bar =12. 26 and s=5. 086

Z=(12. 26-13)/. 7193=-1. 03

Decision Rule: The determined test figure of -1. 03 would not fall in the rejection area of Z< -1. 645, therefore I would not reject the null speculation and state there is insufficient evidence to indicate mu< 13.

Interpretation of Results and Conclusion

p-value=. 152

. 152>. 05

Since the p=value of. 152 is usually greater than the importance level of. 05, I would certainly not reject the null hypothesis at 5% level.

95% CI=(10. eight hundred fifty, 13. 670)- I am 95% self-confident that the common number of years lived in current house falls among 10. eighty-five and 13. 67 years.

Minitab Result

One-Sample Z

Evaluation of mu = 13 vs < 13

The assumed standard deviation = 5. 086...

References: Benson, P. G., McClave, T. T., & Sincich, Big t. (2011). Stats for Business and Economics (11th ed. ). Boston, MOTHER: Prentice Corridor.