# Your Manager Has Speculated the Following

Introduction

Your manager has speculated the following:

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

Null hypothesis is that mean annual income is equal to 50000, and alternative hypothesis is that mean annual income is less than 50000.

H0: µ = 50

Ha: µ < 50

As the population standard deviation is unknown, we’ll use 1-sample t-test.

x-bar = 43.48, s = 14.55, n = 50[O1]

t-statistic = (43.48 - 50)/(14.55/sqrt(50)) = -3.169

df = 50 - 1 = 49

Rejection region: t < -1.677

As test statistic lies in rejection region, H0 is rejected.

There is enough evidence that mean annual income is less than \$50,000.[O2]

P-value = 0.0013

P-value means if actual population mean were equal to \$50,000, then probability of observing this difference was only 0.0013

MINITAB Output for Hypothesis Testing

One-Sample T: INCOME(\$1000)

Test of mu = 50 vs < 50

95% Upper

Variable N Mean StDev SE Mean Bound T P

INCOME(\$1000) 50 43.48 14.55 2.06 46.93 -3.17 0.001

.

95% confidence interval:

Margin of error = 2.01*14.55/sqrt(50) = 4.14

Lower limit = 43.48 - 4.14 = 39.34

Upper limit = 43.48 + 4.14 = 47.62

So, calculated 95% confidence interval for mean is (39.34, 47.62).

This means if we calculated the intervals using other similar samples, then the actual population mean would lie inside the intervals 95% of times. This roughly means, we’re 95% confident that actual population mean is between 40.03 to 46.93.[O3]

MINITAB Output for Confidence Interval

Test of mu = 50 vs not = 50

Variable N Mean StDev SE Mean 95% CI T P

INCOME(\$1000) 50 43.48 14.55 2.06 (39.34, 47.62) -3.17 0.003

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

Null hypothesis is that true proportion is equal to 40% while alternative hypothesis is that true proportion is greater than 40%.

H0: p = 0.40

Ha: p > 0.40

We’ll use 1- proportion z-test.

p-hat = 21/50 = 0.42, n=50

z-statistic = (0.42 - 0.40) / sqrt(0.40*(1-0.40)/50) = 0.289

Rejection region: z > 1.645

As z-statistic doesn’t lie in rejection region, we fail to reject H0.

There isn’t sufficient evidence that true population proportion of customers who live in an urban area exceeds 40%.[O4]

P-value = 0.386

This means if actual population proportion were 40%, then the probability of observing this difference was 0.386.

MINITAB Output for Hypothesis Testing

Test and CI for One Proportion: LOCATION_2

Test of p = 0.4 vs p > 0.4

Event = Urban

95% Lower

Variable X N Sample p Bound Z-Value P-Value

LOCATION_2 21 50 0.420000 0.305190 0.29 0.386

Using the normal approximation.

95% confidence interval:

Margin of error = 1.96*sqrt(0.42*(1-0.42)/50) = 0.137

Lower limit = 0.42 - 0.137 = 0.283

Upper limit = 0.42 + 0.137 = 0.557

So, 95% confidence interval for population proportion is (0.283, 0.557). This roughly means, we’re 95% confident that true population proportion of urban area customers is between 0.283 to 0.557.

MINITAB Output for Confidence Interval

Test of p = 0.4 vs p not = 0.4

Event = Urban

Variable X N Sample p 95% CI Z-Value P-Value

LOCATION_2 21 50 0.420000 (0.283195, 0.556805) 0.29 0.773

Using the normal approximation.

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

Null hypothesis is that mean number of years live in current home is 13 years, while alternative hypothesis is that mean is less than 13.

H0: µ = 13

Ha: µ < 13

As population standard deviation is unknown, we’ll use 1-sample t-test.

x-bar = 12.380, s = 5.103, n = 50

t-statistic = (12.380 - 13) / (5.103/sqrt(50)) = -0.859

df = 50 - 1 = 49

Rejection region: t < -1.677

As t-statistic doesn’t lie in rejection region, we fail to reject H0.

There is not enough evidence that mean number of years lived in current home is less than 13 years.

P-value = 0.197

This means if true population mean was 13 years, then probability of observing this difference was 0.197.

MINITAB Output for Hypothesis Testing

One-Sample T: YEARS

Test of mu = 13 vs < 13

95% Upper

Variable N Mean StDev SE Mean Bound T P

YEARS 50 [O5]12.380 5.103 0.722 13.590 -0.86 0.197

95% Confidence Interval:

Margin of error = 2.01*5.103/sqrt(50) = 1.45

Lower limit = 12.38 - 1.45 = 10.93

Upper limit = 12.38 + 1.45 = 13.83

So, 95% confidence interval for true population mean is (10.93, 13.83). This roughly means we’re 95%

sure that actual population mean is between 10.93 to 13.83 years.[O6]

MINITAB Output for Confidence Interval

Test of mu = 13 vs not = 13

Variable N Mean StDev SE Mean 95% CI T P

YEARS 50 12.380 5.103 0.722 (10.930, 13.830) -0.86 0.394

1. The average (mean) credit balance for suburban customers is more than \$4300.

Null hypothesis is that mean credit balance is equal to \$4300, while alternative hypothesis is that mean credit balance is more than \$4300.

H0: µ = 4300

Ha: µ > 4300

As population standard deviation is unknown, we’ll use 1-sample t-test.

x-bar = 4654, s =770, n = 15[O7]

t-statistic = (4654 - 4300) / (770/sqrt(15)) = 1.781

df = 15 - 1 = 14

Rejection region: t > 1.761

As test statistic lies in rejection region, H0 is rejected.

There is enough evidence that mean credit balance of suburban customers is more than \$4300.

P-value = 0.048

This means if actual mean balance were \$4300, then probability of observing this difference was 0.048.

MINITAB Output for Hypothesis Testing

One-Sample T: CREDIT_SUBURB

Test of mu = 4300 vs > 4300

95%

Lower

Variable N Mean StDev SE Mean Bound T P

CREDIT_SUBURB 15 4654 770 199 4304 1.78 0.048

95% Confidence Interval:

Margin of error = 2.145*770/sqrt(15) = 426.45

Lower limit = 4654 - 426.45 = 4227.55

Upper limit = 4654 + 426.45 = 5080.45

So, 95% confidence interval for the mean credit balance of suburban customers is (4227.55, 5080.45). This roughly means we’re 95% confident that true mean is between 4227.55 and 5080.45[O8]

MINITAB Output for Confidence Interval

Test of mu = 4300 vs not = 4300

Variable N Mean StDev SE Mean 95% CI T P

CREDIT_SUBURB 15 4654 770 199 (4227, 5081) 1.78 0.097

Conclusion?

References?

McClave, J. T., Benson, P. G., & Sincich, T. (2011). Statistics for Business and Economics (11th ed.). Upper Saddle River, NY: Prentice Hall Publishing.

Minitab

[O1]Sample size is greater than 30 therefore by central limit theorem we should use z-test

[O2]type of error?

[O3] type of error?

[O4] type of error?

[O5]Again since n>30 we should use z-test

[O6]type of error?

[O7]Since n<30 are use t-test

[O8]type of error?