Part B Course Project Math 533
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=-3.024 2.070 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
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=-3.024 2.070 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
References: Benson, P. G., McClave, J. T., & Sincich, T. (2011). Statistics for Business and Economics (11th ed.). Boston, MA: Prentice Hall.