Chapter 6: Statistical Inference

Chapter 6: Statistical Inference

Chapter 6: Statistical Inference

1.  The population mean is a fixed number not a random variable, therefore it is incorrect to state that it has a probability to lie in some range of values. It is correct to say that you are “95% confident that the population mean lies between -5 and 5” since this refers to the method by which you are estimating the confidence interval.

2.  False. Accepting the null hypothesis does not mean that the null hypothesis is true. It means that there is insuffient evidence to reject it, accepting the alternative hypothesis.

3.  False. Rejecting the null hypothesis means that the probability of error associated with incorrectly rejecting the null hypothesis is low enough to allow you to accept the alternative hypothesis. However, the null hypothesis might see be true and your decision might still be in error.

a. 

b. 

5. 

a.  Let m= mean price of Civics in San Francisco. The two hypotheses are:

H0 : m = 8500

Ha : m ¹ 8500

We are assuming that the distribution of Honda Civics is normal.

b.  The alternative is two-sided, because the hypothesis is that m differs from the national average, in other words either greater than or less than.

c.  The acceptance region for the null hypothesis covers the range

where s=600, n=9, m=8500, and a=0.05. Finally, use Excel’s NORMSINV function to calculate the value of z1–a/2=NORMSINV(0.975)=1.96. The acceptance region is therefore

Since the acceptance region does not include the sample mean of 9000, we reject the null hypothesis and accept the alternative that the mean price of Civics in San Francisco is not equal to $8500.

d.  One way to perform this test is with a one-sample t test. The t statistic is :

where , m=8500, s=600, and n=10. So that

Using Excel’s TDIST function, TDIST(2.635,9,2)=0.0271. Hence, the probability of the t statistic is 2.71%, and at the 5% significance level, the price difference is significant.

a.  The unknown population mean of the American speakers is μ1, the population mean for the imported speakers is μ2. The null and alternate hypothesis are:

H0: μ1 – μ2 = 0

Ha: μ1 – μ2 ≠ 0

b.  Call the sample average of the American speakers x1, and the sample average of the imported speakers x2. The unknown population mean of the American speakers is m1, the population mean for the imported speakers is m2To test whether these averages indicate that the types of speakers are different in price, use the t-test:

The first group, the American-made speakers, has a sample size, n1 = 10 and a standard deviation, s1 =5. The second group, the imported speakers has a sample size, n2 , equal to 5 with a standard deviation, s2 = 4. Therefore, the pooled standard deviation, s, is:

Under the null hypothesis, the value of the t statistic is:

This t statistic follows the t distribution with 13 degrees of freedom under the null hypothesis. Using Excel’s TDIST function we can calculate the probability of a t distributed random variable having such an extreme value. The form of the Excel function is TDIST(1.936,13,2) = 0.0749. The difference between the two speakers is not significant at the 5% level and we cannot reject the hypothesis of equal population means for the two speaker groups.

c.  It is a significant difference at the 10% level, so at this lower level of statistical significance, we can reject the hypothesis of equal means.

7.  Given the distribution of the t-statistic, we know that:

Multiplying the inequalities by yields:

and subtracing from each term of this inequality, we get:

We multiply each term by –1 (changing the signs and the direction of the inequalities):

and finally, we rearrange the terms and inequalities to arrive at the final form of the confidence interval:

So, the upper and lower confidence limits for μ are:

a.  Let μ1 = mean number of beds for non-rural homes and μ2 = mean number of beds in the rural homes. The null and alternative hypothese are:

H0: μ1 – μ2 = 0

Ha: μ1 – μ2 ≠ 0

b.  The results of the two-sample t-test are:

Descriptive Statistics
N / Mean / Std. Dev / Std. Err
Beds / Location = "Non-rural" / 18 / 111.39 / 43.568 / 10.269
Location = "Rural" / 34 / 83.68 / 36.436 / 6.249
t test Analysis (pooled)
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
Beds / 27.71 / 11.370 / 2.437 / 50.00 / 0.018 / 4.87 / 50.55
t test Analysis (unpooled)
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
Beds / 27.71 / 12.021 / 2.305 / 29.81 / 0.028 / 3.13 / 52.30
Equality of Variance Tests / F-test / Barlett / Levene
0.370 / 0.395 / 0.808

c.  The distribution for the two location types appears as follows:

The standard deviation of the two samples is pretty close (43.568 and 36.436), which would lead us to believe that a pooled estimate of the standard deviation is the way to go. Moreover, the results of the F-test, Bartlett test, and Levene test, support this conclusion. However there is an outlier in both samples, which may cause us to doubt the validity of using the t-test in this situation.

d.  In the Mann-Whitney test, we choose between the following hypotheses:

H0: The median number of beds in the rural nursing homes = median number of beds in the non-rural homes.

Ha: The median number of beds are not equal.

The results of the Mann-Whitney test are:

N / Minimum / 1st Quartile / Median / 3rd Quartile / Maximum
Beds / "Non-rural" / 18 / 60.00 / 81.75 / 120.00 / 120.00 / 244.00
"Rural" / 34 / 25.00 / 59.25 / 80.00 / 106.50 / 221.00
Mann-Whitney Rank Analysis
Median Diff. / Rank Sum1 / Rank Sum2 / p-value / lower 95% / upper 95%
Beds / 26.50 / 613.5 / 764.5 / 0.009 / 4.00 / 52.00

Based on the results of the text, we reject the null hypothesis with a p-value of 0.009 and accept the alternative hypothesis that there are more beds in non-rural homes.

e.  Whether we use the results of the t-tests or the Mann-Whitney test, the conclusion is the same: there are significantly more beds in non-rural homes. The confidence intervals and estimates of the difference between the locations are also very similar.

a.  The first few values of the Days_Beds variable are:

Beds / Medical Days / Total Days / Revenue / Salaries / Expenses / Location / Days Beds
244 / 128 / 385 / 23521 / 5230 / 5334 / Non-rural / 1.578
59 / 155 / 203 / 9160 / 2459 / 493 / Rural / 3.441
120 / 281 / 392 / 21900 / 6304 / 6115 / Non-rural / 3.267
120 / 291 / 419 / 22354 / 6590 / 6346 / Non-rural / 3.492
120 / 238 / 363 / 17421 / 5362 / 6225 / Non-rural / 3.025
65 / 180 / 234 / 10531 / 3622 / 449 / Rural / 3.600
120 / 306 / 372 / 22147 / 4406 / 4998 / Rural / 3.100
90 / 214 / 305 / 14025 / 4173 / 966 / Rural / 3.389
96 / 155 / 169 / 8812 / 1955 / 1260 / Non-rural / 1.760
120 / 133 / 188 / 11729 / 3224 / 6442 / Rural / 1.567

b.  First use a two sample t-test. Let μ1 = mean of the Days_Beds variable for non-rural homes and μ2 = mean of the Days_Beds variable for rural homes. The null and alternative hypotheses are:

H0: μ1 – μ2 = 0

Ha: μ1 – μ2 ≠ 0

Another approach is to use the Mann-Whitney test to evaluate the hypotheses:

H0: The median Days_Beds value in the rural nursing homes = median Days_Beds value in the non-rural homes.

Ha: The median Days_Beds values are not equal.

The results of the two tests are:

Descriptive Statistics
N / Mean / Std. Dev / Std. Err
Days Beds / Location = "Non-rural" / 18 / 3.01505 / 0.576282 / 0.135831
Location = "Rural" / 34 / 3.11144 / 0.637401 / 0.109313
t test Analysis
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
Days Beds / -0.09640 / 0.179938 / -0.536 / 50.00 / 0.595 / -0.45781 / 0.26502
Equality of Variance Tests
F-Test / Bartlett / Levene
0.673 / 0.641 / 0.688
Descriptive Statistics
N / Minimum / 1st Quartile / Median / 3rd Quartile / Maximum
Days Beds / Location = "Non-rural" / 18 / 1.578 / 2.775 / 3.222 / 3.458 / 3.550
Location = "Rural" / 34 / 1.567 / 2.818 / 3.279 / 3.545 / 4.699
Mann-Whitney Rank Analysis
Median Diff. / Rank Sum1 / Rank Sum2 / p-value / lower 95% / upper 95%
Days Beds / -0.061 / 435.500 / 942.500 / 0.419 / -0.322 / 0.165

c.  The distribution of the Days_Beds variable appears as follows:

d.  We fail to reject the null hypothesis under either test. There is no indication that the beds are being utilized at different rates.

a.  Let μ1 = mean draft number in the first half of the year and μ2 = mean draft number in the second half of the year. The null and alternative hypotheses are:

H0: μ1 – μ2 = 0

Ha: μ1 – μ2 ≠ 0

b.  There is not a significant statistic for rejecting the hypothesis that the standard deviation of the two samples are equal, so we'll use the pooled estimate. The result of the two-sample t-test is:

Descriptive Statistics
N / Mean / Std. Dev / Std. Err
Number / Half = 1 / 182 / 206.38 / 106.149 / 7.868
Half = 2 / 184 / 160.92 / 100.757 / 7.428
t test Analysis
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
Number / 45.46 / 10.817 / 4.202 / 364.00 / 0.000 / 24.18 / 66.73
Equality of Variance Tests
F-Test / Bartlett / Levene
0.482 / 0.483 / 0.380

c.  The distribution of the two samples appears as:

The distribution resembles a Uniform distribution, however since the t-test is robust to problems with non-Normality, it can still be used here.

d.  The 95% confidence intervals are:

N / Mean / Std. Dev / Std. Err / lower 95% / upper 95%
Half = 1 / 182 / 206.38 / 106.149 / 7.868 / 190.85 / 221.9
Half = 2 / 184 / 160.92 / 100.757 / 7.428 / 146.27 / 175.58

e.  A draft number selected for a person born in the first half of the year is, on average, 45.46 points higher than a draft number for a person whose birthday falls in the second half of the year. This is a statistically significant difference. The ramification of this result is that the assignment of draft numbers is not truly random and that people born in the second half of the year are more likely to receive low draft numbers, and hence are more likely to be drafted.

11. 

a.  Let μ1 = mean female salary and μ2 = mean male salary. The null and alternative hypotheses are:

H0: μ1 – μ2 = 0

Ha: μ1 – μ2 ≠ 0

We'll use a significance level of 5% for this test.

b.  The unpooled and pooled t-tests results are:

Descriptive Statistics
N / Mean / Std. Dev / Std. Err
Salary / Sex = "F" / 37 / 27,027.35 / 5,478.231 / 900.616
Sex = "M" / 44 / 33,004.91 / 5,831.991 / 879.206
unpooled t test Analysis
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
Salary / -5,977.56 / 1,258.615 / -4.749 / 78.00 / 0.000 / -8,483.27 / -3,471.85
pooled t test Analysis
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
Salary / -5,977.56 / 1,265.517 / -4.723 / 79.00 / 0.000 / -8,496.51 / -3,458.61
Equality of Variance Tests
F-Test / Bartlett / Levene
0.705 / 0.698 / 0.000

The conclusions of the two tests are the same, suggesting that there is a statistically significant difference in salary between the male and female professors. A histogram of the difference appears as follows:

c.  There are only enough data for the assistant professor and instructor groups. The unpooled and pooled t-test results are:

Descriptive Statistics
N / Mean / Std. Dev / Std. Err
asst prof / Sex = "F" / 17 / 28,274.18 / 6,598.744 / 1,600.430
Sex = "M" / 15 / 31,202.33 / 5,027.220 / 1,298.023
instructor / Sex = "F" / 20 / 25,967.54 / 4,197.807 / 938.658
Sex = "M" / 17 / 30,960.41 / 5,829.235 / 1,413.797
unpooled t test Analysis
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
asst prof / -2,928.16 / 2,060.641 / -1.421 / 29.42 / 0.166 / -7,142.64 / 1,286.33
instructor / -4,992.87 / 1,697.027 / -2.942 / 28.54 / 0.006 / -8,469.08 / -1,516.66
pooled t test Analysis
Mean Diff. / Std. Err. / t / df / p-value / lower 95% / upper 95%
asst prof / -2,928.16 / 2,096.262 / -1.397 / 30.00 / 0.173 / -7,209.29 / 1,352.98
instructor / -4,992.87 / 1,652.707 / -3.021 / 35.00 / 0.005 / -8,348.05 / -1,637.69
Equality of Variance Tests
F-Test / Bartlett / Levene
asst prof / 0.312 / 0.307 / 0.173
instructor / 0.173 / 0.177 / 0.005

The conclusion from the pooled and unpooled t-tests is the same: there is a significant difference for instructors, but not one for assistant professors.

d.  There is some evidence that the university has underpaid its female faculty members. The overall significance level for all ranks is less than 0.0001. When testing for individual ranks, there is statistical significance only for the instructors. There is no significant difference for assistant professors and not enough evidence for other teaching ranks.