Chapter 7 - Statistical Inference: Population mean and proportion

Question 1 If last years mean income of a small town was $25,300 and this year a survey of 40 random household in that town showed a mean income of $29,400 and a standard deviation of $6,325.

(a) Find a 95% confidence interval for the mean income for the town.

(b) Based on answer in part (a) could you conclude that the mean income increased from last year to this year?

Choose Test Name: (Test for population mean, Large sample Size)
 

Enter sample and population parameters: = 6325/6.325 = 1000  Sample size, n = 40 (Large)  =  $6,325  = $29,400  and  = 25,300  (the population mean given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    >
(2) Choose level of Significance,  = 5% or 0.05 Write  = 0.95		Enter degrees of freedom, d.f.  n = 40
(3) Write and Compute the Test Statistics ( z-score):  = (29400 - 25,300) / (6,325 / 6.3246) =  4.1
(4) Select Critical Region of Test Criterion:  z-score for   z0.025 = -1.95  and z0.975 = 1.96 The 95% confidence interval is  + (z0.025)  to  +  (z0.975 )= 29,400 - 1.96(1000) to 29,400 + 1.96(1000) (a) So 95% confidence interval  is $27,439.86 to $31,360.14  (The critical region is outside this interval)
(5) Make Decision: Since 4.1 is > than 1.96 and 25,300 is not within the confidence interval we reject the H0    that   =  (b) Therefore this year's mean of $29,400 is significantly larger than $25,300

Question 2 The noise level tested at an average decibel of 65 with a standard deviation of 6 from a sample of 40 cars driving on the Columbus Bridge.

(a) Find a 90% confidence interval for the mean decibel levels of vehicles driving on the bridge.

(b) Is the noise level significantly lower that 80 decibels?

(c) What is the maximum error of the estimate at the 90% level?
 

Choose Test Name: (Test for population mean, Large sample Size)
 

Enter sample and population parameters: = 6/6.325 = 0.9487 Sample size, n = 40 (Large)  =  6  = 65  and  = 80  (a reference given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    <
(2) Choose level of Significance,  = 10% or 0.10 Write  = 0.90		Enter degrees of freedom, d.f.  n = 40
(3) Write and Compute the Test Statistics ( z-score):  = (65 - 80) / 0.9487 =  -15.8111
(4) Select Critical Region of Test Criterion:  z-score for   z0.05 = -1.65  and z0.95 = 1.65 The 95% confidence interval is  + (z0.05)  to  +  (z0.95 )= 65 - 1.65(0.9487) to 65 + 1.65(0.9487) (a) So 95% confidence interval  is 63.43 to 66.57  (The critical region is outside this interval)
(5) Make Decision: Since -15.8111 is < than -1.65 and 80 is not within the confidence interval we reject the H0    that   =  (b) Therefore the noise level of 65 decibles  is significantly lower than 80 decibles.

(c) The maximum error of the estimate, E = z() = 1.65(0.9487) = 1.57 decibles

Question 3 A study of the dietary calories from fat intakes of 33 males living on a small Caribbean island gave a sample mean of 23.9% and a sample standard deviation of 4.6% calories from fats.

(a) Find a 95% confidence interval for the mean of the population from the men selected.

(b) Is the percent calories from fat equal to 25%?

(c) What is the maximum error of the estimate for the population mean?

Choose Test Name: (Test for population mean, Large sample Size)
 

Enter sample and population parameters: = 4.6/5.831 = 0.7889 Sample size, n = 33 (Large)  = 4.6  = 23.9  and  = 25 (a reference given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    <
(2) Choose level of Significance,  = 5% or 0.05 Write  = 0.95		Enter degrees of freedom, d.f. n = 34
(3) Write and Compute the Test Statistics ( z-score):  = (23.9 - 25) / 0.7889 =  -1.3943
(4) Select Critical Region of Test Criterion:  z-score for    z0.025 = -1.95  and z0.975 = 1.96 The 95% confidence interval is  + (z0.025)  to  +  (z0.975 )= 23.9 - 1.96(0.7889) to 23.9 + 1.96(0.7889) (a) So 95% confidence interval  is 22.35 to 25.45  (The critical region is outside this interval)
(5) Make Decision: Since -1.3943 is between -1.65 and 1.65 and 25 is within the confidence interval we acceptthe H0  that   =  (We Donot reject H0) (b) Therefore the caloriic intake from fat is equal to 25%.

(c) The maximum error of the estimate, E = z() = 1.96(0.7889) = 1.55%

Question 4  How many household in a small town must be randomly selected to estimate the dollar amount spend on animal products per week within $3 of its true value with a 90% confidence if the standard deviation from past studies was $25?

:   z (90%) or alpha = 0.10 and alpha / 2 = 0.05  so z_0.05 = 1.65, s = 25 and E = 3, so n = z(s)/E = (1.65)(25)/3 = 13.75

Question 5 A speculator believes that the average depth of finding water for well installation on newly developed lands in Duchess County is less that 500 feet. If a sample of 32 wells throughout the country gave a mean depth of 486 feet and a standard deviation of 53 feet. At the 1% confidence level is the speculator's assumption correct?

Choose Test Name: (Test for population mean, Large sample Size)
 

Enter sample and population parameters: = 53/5.6568 = 9.3693 Sample size, n = 32 (Large)  = 53  = 486  and  = 500 (a reference given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    <
(2) Choose level of Significance,  = 1% or 0.01 Write  = 0.99		Enter degrees of freedom, d.f. n = 32
(3) Write and Compute the Test Statistics ( z-score):  = (486 - 500) / 9.3693 =  -1.4942
(4) Select Critical Region of Test Criterion:  z-score for    z0.01 = -2.33 a lower tail test Accept null hypothesis if
(5) Make Decision: Since -1.4942  is not in the critical region we accept the H0  that   =   (We Donot reject H0) Therefore the mean depth of wells are close to 500 feet.

Question 6  A doctor recorded the recovery times in days for 38 patients from a new procedure and hope that the recovery time would be less that other procedures of 6.5 days. If the mean recovery time for patients from the sample is 5.84 days and the standard deviation is 2.41 days. Does this data support the doctors hypothesis?

(a) Use the classical methods of confidence interval to draw your conclusion

(b) use the P-value approach

Choose Test Name: (Test for population mean, Large sample Size)
 

Enter sample and population parameters: = 2.41/6.1644 = 0.391 Sample size, n = 38 (Large)  =  2.41  = 5.84   and  = 6.5 (a reference given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    <
(2) Choose level of Significance,  = 5% or 0.05 Write  = 0.95		Enter degrees of freedom, d.f.  n = 38
(3) Write and Compute the Test Statistics ( z-score):  = (5.84 - 6.5) / 0.391 =  -1.688
(4) Select Critical Region of Test Criterion:  z-score for   z0.05 = -1.65  Accept null hypothesis if
(5) Make Decision: Since -1.688 if < 01.65 we reject the H0    that   =  (b) Therefore the reciver time is less than 6.5 days, on average.

Question 7 On a mathematics final exam the following sample distribution was obtained.
Is the average results from the final exam the same as last years average of 75% at the 5 % level of significance?
 

Grade	50	60	70	80	90
Number of Students	2	4	6	4	2

First Find the mean and standard deviation from the sample distribution
 

Group Mean = 70 and standard deviation  = 11.8818

Choose Test Name: (Test for population mean, small sample Size)
 

Enter sample and population parameters: = 11.8818/4.2426 = 2.8006 Sample size, n = 18 (small)  =  11.8818  = 70   and  = 70 (last years - a reference given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    <
(2) Choose level of Significance,  = 5% or 0.05 Write  = 0.95		Enter degrees of freedom, d.f. = n - 1 = 18 - 1 = 17
(3) Write and Compute the Test Statistics ( t-test):  = (70 - 75) / 2.8006 =  -1.7853
(4) Select Critical Region of Test Criterion:  t-statistics for   t0.05 = -1.7396  Accept null hypothesis if (Lower-Tailed Test ():)    Accept H0 if
(5) Make Decision: Since -1.7853 is < -1.7396   we reject the H0    that   =  (b) Therefore this years score is less than last year's.

Question 8. A researcher suspected that smokers between the age of 40 and 45 who had developed chronic bronchitis
had smoked for more than 20 years. A sample of 10 such patients gave the number of years that the smoked which is given in the table below.
At the 1% significance level is there enough evidence to justify the researcher's suspicion?
 

22 21 19 25 24 26 23 23 21 22

First compute the mean and standard deviation:
Mean = 22.6 and standard deviation = 2.0656

Choose Test Name: (Test for population mean, small sample Size)
 

Enter sample and population parameters: = 2.0656 /3.1623 = 0.6532 Sample size, n = 10 (small)  =  2.0656 = 22.6   and  = 20 (a reference given above)
(1) Select Null Hypothesis: H0   =		Alternate Hypothesis: Ha    <
(2) Choose level of Significance,  = 1% or 0.01 Write  = 0.99		Enter degrees of freedom, d.f. =  n- 1 = 10 - 1 = 9
(3) Write and Compute the Test Statistics ( t-test):  = (22.6 - 20) / 0.6532 =  3.98
(4) Select Critical Region of Test Criterion:  t-statistics for   t0.01 = 2.8214  Accept null hypothesis if (Upper-Tailed Test ():)     Accept H0 if
(5) Make Decision: Since 3.98 > 2.8214   we reject the H0    that   =  (b) Therefore smokers between the age of 40 and 45 who had developed chronic bronchitis had smoked for more than 20 years So the researcher claim is true.

Question 9 A study of 298 men treated for prostate cancer at an early stage had 271 recovered within 5 years.

(a) Find a 95% confidence interval for the proportion of men who recovered within 5 years so treated.

(b) Could you say that 95% of the men recovered at the 95% significance level?

Choose Test Name: (Test for population proportion, Large sample Size)
 

Enter sample and population parameters: s == 0.01662 Sample size, n = 298 (Large),  p = 271/298 = 0.9094   and  = 0.95  (a reference given above)
(1) Select Null Hypothesis: H0		Alternate Hypothesis:
(2) Choose level of Significance,  = 5% or 0.05 Write  = 0.95		Enter degrees of freedom, d.f.  n = 298
(3) Write and Compute the Test Statistics ( z-score):  = (0.9094 - 0.95) / 0.01662 =  -2.454
(4) Select Critical Region of Test Criterion:  z-score for   z0.025 = -1.96  and z0.975 = 1.96 The 95% confidence interval is p + (z0.025)  to p +  (z0.975 )= 0.9094 - 1.96(0.01662)  to 0.9094  + 1.96(0.01662) (a) So 95% confidence interval  is  0.8768 to 0.942   (The critical region is outside this interval) Lower-Tailed Test ():  Accept H0 if
(5) Make Decision: Since 0.95 is outside the CI of   0.8768 to 0.942 and z-stat < z-alpha, -2.452 < -1.96 we reject the H0 (b) Therefore we cannot say that 95% of the men who recover are of the same population.

Question 10 In a poll of customers who buy a service 383 of 935 said that they were not satisfied with service the were it was.
Find a 95% confidence interval of the number of customers unsatisfied with service.

Choose Test Name: (Test for population proportion, Large sample Size)
 

Enter sample and population parameters: s == 0.01608 Sample size, n = 935 (Large),  p = 383/935 = 0.4096
(1) Select Null Hypothesis: H0    not applicable		Alternate Hypothesis:  not applicable
(2) Choose level of Significance,  = 5% or 0.05 Write  = 0.95		Enter degrees of freedom, d.f.  n = 935
(3) Write and Compute the Test Statistics ( z-score):   not applicable since there is no reference
(4) Select Critical Region of Test Criterion:  z-score for   z0.025 = -1.96  and z0.975 = 1.96 The 95% confidence interval is p + (z0.025)  to p +  (z0.975 )= 0.4096 - 1.96(0.01608)  to 0.4096 + 1.96(0.01608) Answer:  So 95% confidence interval  is  0.3781 to 0.4411  (The critical region is outside this interval)
(5) Make Decision: the CI of   0.3781 to 0.4411  any claim ouside this interval would cause a rejection of H0