Introduction
Your manager has speculated the following:
- 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
- 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.
- 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
- 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?