Part B Course Project Math 533

Topics: Statistical hypothesis testing, Statistics, Null hypothesis Pages: 8 (1859 words) Published: August 24, 2013
Course Project Part B

a.
the average (mean) annual income was less than $50,000

Null and Alternative Hypothesis

H0: mu= 50 (in thousands)
Ha: mu<50 (in thousands)

Level of Significance
Level of Significance = .05

Test Statistic, Critical Value, and Decision Rule

Since alpha = .05, z<-1.645, which is lower tailed
Rejection region is, z<-1.645

Calculate test statistic, x-bar=43.74 and s=14.64
Z=(43.74-50)/2.070=-3.0242.070 is calculated by: s/sq-root of n

Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.

Interpretation of Results and Conclusion

p-value= .001
.001<.05

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

95% CI=(39.68, 47.80)- I am 95% confident that the true mean income lies between $39,680 and $47,800.
Minitab Output:
One-Sample Z

Test of mu = 50 vs < 50
The assumed standard deviation = 14.64

95% Upper
N Mean SE Mean Bound Z P
50 43.74 2.07 47.15 -3.02 0.001

b. the true population proportion of customers who live in an urban area exceeds 40%

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

Null and Alternative Hypothesis

H0: p=.40
Ha: p>.40

Level of Significance

Level of Significance= .05

Test Statistic, Critical Value, and Decision Rule

Since alpha= .05, z>1.645, which is upper tailed
Rejection region is, z>1.645

To conduct a large sample z-test, I must first determine if the sample size is large enough. nPo= 50(.40)= 20 and 50(1-.40)=30
Both are larger than 15, so we conclude that the sample size is large enough to conduct the large sample z test.

Z=(.44-.40)/.06928=.5774.06928 is calculated by sq-root ((.4)(.6))/50)=.06928

Decision Rule: The calculated test statistic of .5774 does not fall in the rejection region of z>1.645, therefore I would not reject H0. There is insufficient evidence to conclude the true population of customers who live in urban communities is greater than 40%.

Interpretation of Results and Conclusions

p-value= .282
.282>.05

Because the p-value of .282 is greater than the significance level of .05, I will not reject the null.

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

Minitab Output
Test and CI for One Proportion

Test of p = 0.4 vs p > 0.4

95% Lower
Sample X N Sample p Bound Z-Value P-Value
1 22 50 0.440000 0.324532 0.58 0.282

Using the normal approximation.

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

Null and Alternative Hypothesis

H0: mu=13
Ha: mu<13

Level of Significance

Level of Significance= .05

Test Statistic, Critical Value, and Decision Rule

Since alpha= .05, z<-1.645, which is lower tailed.
Rejection region is z<-1.645

Calculate the test statistic, x-bar =12.26 and s=5.086
Z=(12.26-13)/.7193=-1.03

Decision Rule: The calculated test statistic of -1.03 does not fall in the rejection region of Z<-1.645, therefore I would not reject the null hypothesis and say there is insufficient evidence to indicate mu<13.

Interpretation of Results and Conclusion

p-value=.152
.152>.05

Because the p=value of .152 is greater than the significance level of .05, I would not reject the null hypothesis at 5% level.

95% CI=(10.850, 13.670)- I am 95% confident that the average number of years lived in current home falls between 10.85 and 13.67 years.

Minitab Output

One-Sample Z

Test of mu = 13 vs < 13
The assumed standard deviation = 5.086...

References: Benson, P. G., McClave, J. T., & Sincich, T. (2011). Statistics for Business and Economics (11th ed.). Boston, MA: Prentice Hall.
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