Project Part B

Answer the following questions using the data in a the AJ case file (the data set you already have in your course file)

A. The Manager of AJ believe the average (mean) annual income of customers was less than

$50,000. Using the data set annual income, is there convincing evidence that the true mean

annual salary is less than $50,000 ? Use =.05.

1. Null hypothesis (Ho): The average(mean) annual income was ≤$50,000

Ho: µ

Ho: µ =50

2. Alternative (research) hypothesis (Ha): The average (mean) annual income < $50,000

Ha: µ

H1: µ <50

3. Test statistic:

Z= (43.74-50)/7.0711= 2.08, because =-2.07,because =

14.64/ = 7.07107

4. Rejection region:

For a= 0.5 and “<” in the Ha, I found that z= -1.645, so the “Rejection Region” would be z<-1.645

5. Assumptions:

If the p-value is less than alpha, reject the null hypothesis and accept the alternative hypothesis, at the given alpha. When you look at the calculated test statistics results you can see that both the test statistic and the p-value methods have the same reject or not reject results.

Because the p-value = 0.001 is less than alpha = 0.05: we reject the null hypothesis H0: μ=50 and we accept the alternative hypothesis Ha: μ<50, at α=0.05.

6. Experiment and calculation of test statistic: ( Use MINITAB OUTPUT ANNUAL INCOME)

Descriptive Statistics: Income ($1000)

Variable Mean StDev Minimum Maximum

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Income ($1000) 43.74 14.64 21.00 67.00

One-Sample Z

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

95% Upper N Mean SE Mean Bound Z P50 43.74 2.07 47.15 -3.02 0.001

7.Conclusion:

My calculated test statistic of -2.07 falls in the rejection region of z < -0.1645, therefore, I would reject the null hypothesis and say there is sufficient evidence to indicate u<50 or $50,000.

The assumed standard deviation = 14.64

Variable N Mean StDev SE Mean 95% CIIncome ($1000) 50 43.74 14.64 2.07 (39.68, 47.80)

According to the confidence interval, we are 95% confident that the true mean income lies between $39,680 and $47,800.

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

1. Null hypothesis (Ho):

Ho: = 0.40

2. Alternative (research) hypothesis (Ha):

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Ha: p>0.40

3. Test statistic: (MINITAB OUTPUT)

Number of Events (Urban Customers)

Location Count Percent Rural 13 26.00Suburban 15 30.00 Urban 22 44.00 N= 50

4. Rejection region:

Z= (0.44 – 0.400)/ 0.69282= 0.58 where s phat= sqrt (((0.40) (0.60))/50= 0.069282

This is a one tailed (upper or right since HA has “>”). Our rejection regions would be z > 1.645.

0.58 is not greater than 1.645 (and is not in the rejection regions) so we would not Reject the Ho.

5. Assumptions:

If the p-value is less than alpha, reject the null hypothesis and accept the alternative hypothesis, at the given alpha. When you look at the calculated test statistics results you can see that both the test statistic and the p-value methods have the same reject or not reject results.

Because the p-value = 0.282 is more than alpha = 0.05: we do not reject the null hypothesis H0: μ=40 and we do not accept the alternative hypothesis

Ha: μ<40, at α=.05.

Since we are not rejecting the Ho, we are saying there is insufficient evidence to conclude the true population of customers who live in the Suburban location is greater than 40%.

6. Experiment and calculation of test statistic: (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

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22 50 0.440000 0.324532 0.58 0.282

Sample X N Sample p 95% CI 1 22 50 0.440000 (0.302411, 0.577589)

7. Conclusion:

Sample X N Sample p 95% CI Z-Value P-Value1 22 50 0.440000 (0.302411, 0.577589) 0.58 0.564

According to the confidence interval, we are 95% confident that the mean population lies between 0.302 and 0.577.

C. The AJ manager assumed average (mean) number of years lived in the current home is less than 13 years. Is this assumption correct? Choose alpha. (ii) Additionally, calculate a 95% confidence interval for this tests statistic value.

1. Null hypothesis (Ho):

Ho: u = 13

2. Alternative (research) hypothesis (Ha):

H1: u<13

3. Test statistic:

z= (12.26 -13)/0.7193= -1.03, because = 5.086/ √n (50)= 0.7193

4. Rejection region:

For a = 0005 and “<” in the Ha, I found that z= -1.645, so the “rejection Region would be z < -1.645

5. Assumptions:

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Because the p-value = 0.152 is more than alpha = 0.05: we do not reject the null hypothesis H0: μ=13 and we do not accept the alternative hypothesis Ha: μ<13, at α=.05.

6. Experiment and calculation of test statistic: (Use MINITAB OUTPUT)

One-Sample Z: Years

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

95% UpperVariable N Mean StDev SE Mean Bound Z PYears 50 12.260 5.086 0.719 13.443 -1.03 0.152

7. Conclusion:

According to the confidence interval, we are 95% confident that the average mean of people living

in their current homes lies between 42.33 and 45.15.

C. Is the average (mean) credit balance for suburban customers is more than $4300? Choose your own alpha and test the hypothesis.(ii) Additionally, calculate a 95% confidence interval for this tests statistic value.

1. Null hypothesis (Ho):

Ho: u = 4300

2. Alternative (research) hypothesis (Ha):

H1: u> 4300

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3. Test statistic:

One-Sample Z: Credit Balance ($)

The assumed standard deviation = 932

Variable N Mean StDev SE Mean 95% CICredit Balance($) 50 3970 932 132 (3712, 4229)

4. Rejection region:

For a = .05 and “>” in the Ha, I found z= 1.645, so the Rejection Region would bez > 1.645.

5. Assumptions:

If the p-value is less than alpha, reject the null hypothesis and accept the alternative hypothesis, at the given alpha. When you look at the calculated test statistics results you can see that both the test statistic and the p-value methods have the same reject or not reject results.

Because the p-value = 0.994 is not less than alpha = .05: we do not reject the null hypothesis H0: μ=4300 and we do not accept the alternative hypothesis Ha: μ>4300 at α=.05.

6. Experiment and calculation of test statistic: (MINITAB OUTPUT)

Test of mu = 4300 vs > 4300The assumed standard deviation = 932 95% LowerVariable N Mean StDev SE Mean Bound Z PCredit Balance($) 50 3970 932 132 3754 -2.50 0.994

7. Conclusion:

We are 95% confident that the true mean credit balance lies between $3,712 and $4,229.

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