Comparing the potency of a particular drug: Fresh versus Stored

Data:

```
> # potency readings for Fresh drug
> cFresh = c(10.2, 10.5, 10.3, 10.8, 9.8, 10.6, 10.7, 10.2, 10,
+ 10.6)
>
> # potency readings for Stored drug
> cStored = c(9.8, 9.6, 10.1, 10.2, 10.1, 9.7, 9.5, 9.6, 9.8, 9.9)
>
> # combine them into data frame
> Length = as.data.frame(cbind(cFresh, cStored))
> colnames(Length) = c("Fresh", "Stored")
```

How to access if the mean potency of Fresh drug is different than that of Stored drug?

- how to obtain information on the two means?
- how to summarize information on the two means?
- what is a test statistic for this task?

Standard deviations for each sample

```
> # std dev for potency of Fresh drug
> sd(cFresh)
[1] 0.3233505
>
> # std dev for potency of Stored drug
> sd(cStored)
[1] 0.2406011
```

Question: the population standard deviations are the same?

Strategy to access the difference between two means:

- obtain sample means from the two samples
- take the difference between the two sample means
- normalize the difference (difficult!)
- assess the normalized difference statistically

```
> mean(cFresh)
[1] 10.37
> mean(cStored)
[1] 9.83
> mean(cFresh) - mean(cStored)
[1] 0.54
>
> # sample standard deviation of differences
> sd(cFresh - cStored)
[1] 0.4325634
```

If \(y_1 \sim \mathsf{N}(\mu_1,\sigma_1^2)\), and \(y_2 \sim \mathsf{N}(\mu_2,\sigma_2^2)\) and they are independent, then the difference \(y_1 - y_2\) follows

\[\mathsf{N}(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2)\]

Similarly, the sum \(y_1 + y_2\) follows

\[\mathsf{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)\]

Note the variance term in the above

- Sample for Fresh drug follow \(\mathsf{N}(\mu_1,\sigma_1^2)\)
Sample for Stored drug follow \(\mathsf{N}(\mu_2,\sigma_2^2)\)

Sample mean for Fresh drug: \[\bar{y}_1 \sim \mathsf{N}(\mu_1,\sigma_1^2/n)\]

Sample mean for Stored drug \[\bar{y}_2 \sim \mathsf{N}(\mu_2,\sigma_2^2/n)\]

Further \(d = \bar{y}_1 - \bar{y}_2\) follows \[\mathsf{N}\left(\mu_1 - \mu_2,\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}\right)\]

Assume \(\sigma_1 = \sigma_2\) and set \[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2 -1)s_2^2}{n_1 + n_2 -2}}\] then \(T = \frac{d}{s_{p}}\) follows a Student’s t distribution with degrees of freedom (df) \(n_1 + n_2 -2\)

Assumptions:

- Random sample for Fresh drug follow \(\mathsf{N}(\mu_1,\sigma_1^2)\)
- Random sample for Stored drug follow \(\mathsf{N}(\mu_2,\sigma_2^2)\)
- Two samples are independent
- Assume \(\sigma_1 = \sigma_2\)

The \(100(1- \alpha)\%\) confidence interval (CI) for the difference \(\mu_1 - \mu_2\) is constructed as follows:

Difference between sample means \(d = \bar{y}_1 - \bar{y}_2\)

- Set \(s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2 -1)s_2^2}{n_1 + n_2 -2}}\)
CI: \[ (\bar{y}_1 - \bar{y}_2) \pm t_{\alpha/2} s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \] where \(t_{\alpha/2}\) is the \((1-\alpha/2)\%\) quantile of a Student’s t distribution with \(df = n_1+n_2-2\)

Illustration of CI: the \(95\%\) CI is \([0.2722297,0.8077703]\)

```
> n1 = length(cFresh)
> n2 = length(cStored)
> d = mean(cFresh) - mean(cStored)
> sp = sqrt(((n1 - 1) * (sd(cFresh))^2 + (n2 - 1) * (sd(cStored))^2)/(n1 +
+ n2 - 2))
> cval = qt(0.05/2, df = n1 + n2 - 2, ncp = 0, lower.tail = FALSE)
> cval
[1] 2.100922
> nr = sqrt(1/n1 + 1/n2)
> CI_left = d - cval * sp * nr
> CI_left
[1] 0.2722297
> CI_right = d + cval * sp * nr
> CI_right
[1] 0.8077703
```

Illustration of CI: the \(95\%\) CI is \([0.2722297,0.8077703]\)

```
> tTest = t.test(x = cFresh, y = cStored, alternative = "two.sided",
+ mu = 0, paired = FALSE, var.equal = TRUE, conf.level = 0.95)
> tTest$conf.int
[1] 0.2722297 0.8077703
attr(,"conf.level")
[1] 0.95
```

Hypothesis testing when \(D_0\) is a postulated value related to \(\mu_1 - \mu_2\):

Recall \(d = \bar{y}_1 - \bar{y}_2\) and \[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2 -1)s_2^2}{n_1 + n_2 -2}}\]

Test statistic: \(T = \frac{d - D_0}{s_{p} \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\) follows a Student’s t distribution with \(df= n_1 + n_2 - 2\)

Illustration: \(H_0: \mu_1 - \mu_2 = 0\) vs \(H_0: \mu_1 - \mu_2 \ne 0\)

- \(D_0 = 0\) and \(t_{0.05/2,18} = 2.100922\)
- Reject \(H_0\) if \(|T| \ge t_{\alpha/2,n_1 + n_2 -2}\)

```
> t = (d - 0)/(sp * nr)
> t #value of test stat
[1] 4.236833
> cval #critical value
[1] 2.100922
> t < cval
[1] FALSE
> pval = 2 * pt(t, df = n1 + n2 - 2, ncp = 0, lower.tail = FALSE)
> pval
[1] 0.0004959478
```

Test \(H_0: \mu_1 - \mu_2 = 0\) vs \(H_0: \mu_1 - \mu_2 \ne 0\)

- \(D_0 = 0\) and \(t_{0.05/2,18} = 2.100922\)
- Reject \(H_0\) if \(|T| \ge t_{\alpha/2,n_1 + n_2 -2}\)

```
> t.test(x = cFresh, y = cStored, alternative = "two.sided", mu = 0,
+ paired = FALSE, var.equal = TRUE, conf.level = 0.98)
Two Sample t-test
data: cFresh and cStored
t = 4.2368, df = 18, p-value = 0.0004959
alternative hypothesis: true difference in means is not equal to 0
98 percent confidence interval:
0.2146898 0.8653102
sample estimates:
mean of x mean of y
10.37 9.83
```

Test \(H_0: \mu_1 - \mu_2 \le 0.1\) vs \(H_0: \mu_1 - \mu_2 > 0.1\)

- \(D_0 = 0.1\) and \(t_{0.05,18} = 1.734064\)
- Reject \(H_0\) if \(T \ge t_{\alpha,n_1 + n_2 -2}\)

```
> t.test(x = cFresh, y = cStored, alternative = "greater", mu = 0.1,
+ paired = FALSE, var.equal = TRUE, conf.level = 0.99)
Two Sample t-test
data: cFresh and cStored
t = 3.4522, df = 18, p-value = 0.001421
alternative hypothesis: true difference in means is greater than 0.1
99 percent confidence interval:
0.2146898 Inf
sample estimates:
mean of x mean of y
10.37 9.83
```

Test \(H_0: \mu_1 - \mu_2 \ge 0.5\) vs \(H_0: \mu_1 - \mu_2 < 0.5\)

- \(D_0 = 0.5\) and \(- t_{0.05,18} = -1.734064\)
- Reject \(H_0\) if \(T \le - t_{\alpha,n_1 + n_2 -2}\)

```
> t.test(x = cFresh, y = cStored, alternative = "less", mu = 0.5,
+ paired = FALSE, var.equal = TRUE, conf.level = 0.96)
Two Sample t-test
data: cFresh and cStored
t = 0.31384, df = 18, p-value = 0.6214
alternative hypothesis: true difference in means is less than 0.5
96 percent confidence interval:
-Inf 0.7764699
sample estimates:
mean of x mean of y
10.37 9.83
```

Recall the following:

- Fresh drug: \(n_1 = 10\), \(\bar{y}_1=10.37\), \(s_1 = 0.3233\)
- Stored drug: \(n_2 = 10\), \(\bar{y}_2=9.83\), \(s_2 = 0.2406\)
- Assume unequal variances, i.e., \(\sigma_1^2 \ne \sigma_2^2\)

The \(100(1- \alpha)\%\) confidence interval (CI) for the difference \(\mu_1 - \mu_2\) is constructed as follows:

- Recall \(d = \bar{y}_1 - \bar{y}_2\) and set \(\tilde{s}_p = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\)
- Test statistic: \(\tilde{T} = \frac{d - (\mu_1 - \mu_2)}{\tilde{s}_p}\) is approximated by a Student’s t distribution with df given earlier
- CI: \[(\bar{y}_1 - \bar{y}_2) \pm t_{\alpha/2,\textrm{df}}\tilde{s}_p\]

Illustration on CI

- \(d = \bar{y}_1 - \bar{y}_2 = 10.37 - 9.83 = 0.54\)
- \(\tilde{s}_p = 0.1274\)
- \(df = 16.62774\)
- \(t_{0.025,17} = 2.11\); T table

CI: \(0.54 \pm 2.11 \times 0.1274\), i.e., \([0.271, 0.808]\)

Testing when \(D_0\) is a postulated value related to \(\mu_1 - \mu_2\):

- Recall \(d = \bar{y}_1 - \bar{y}_2\) and set \(\tilde{s}_p = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\)
- Test statistic: \(\tilde{T} = \frac{d - D_0}{\tilde{s}_p}\) approximated by a Student’s t distribution with \[df= \frac{(n_1-1)(n_2-1)}{(1-c)^2(n_1-1)+c^2(n_2-1)}\] and \(c = \frac{s_1^2/n_1}{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\)
- Note: round df to the nearest integer

At Type I error probability \(\alpha=0.05\)

Test \(H_0: \mu_1 - \mu_2 \le 0.2\) vs \(H_a: \mu_1 - \mu_2 > 0.2\)

- \(D_0 = 0.2\)
Reject \(H_0\) if \(\tilde{T} \ge t_{\alpha,df}\)

Obtain value of test statistic

- \(\tilde{s}_p = 0.1274\)
- \(df = 16.62774\)
- \(t = \frac{10.37 - 9.83 - 0.2}{0.1274} = 2.668\)
- \(t_{0.05,17} = 1.74185\); T table

Compare \(t\) with \(1.74185\)

At Type I error probability \(\alpha=0.05\)

Test \(H_0: \mu_1 - \mu_2 = D_0\) vs \(H_a: \mu_1 - \mu_2 \ne D_0\)

- \(D_0 = 0\) and \(t_{0.025,17} = 2.11\)
Reject \(H_0\) if \(|\tilde{T}| \ge t_{\alpha/2,df}\)

At Type I error probability \(\alpha=0.05\)

Test \(H_0: \mu_1 - \mu_2 \ge D_0\) vs \(H_0: \mu_1 - \mu_2 < D_0\)

- \(D_0 = 0.3\) and \(t_{0.05,17} = 1.74185\)
Reject \(H_0\) if \(\tilde{T} \le - t_{\alpha,df}\)

```
> potency = read.table("http://math.wsu.edu/faculty/xchen/stat412/data/t_uev.txt",
+ sep = "\t", header = TRUE)
> class(potency)
[1] "data.frame"
> cFresh = potency$Fresh
> cFresh
[1] 10.2 10.5 10.3 10.8 9.8 10.6 10.7 10.2 10.0 10.6
> cStored = potency$Stored
> cStored
[1] 9.8 9.6 10.1 10.2 10.1 9.7 9.5 9.6 9.8 9.9
```

At Type I error probability \(\alpha=0.05\)

- Test \(H_0: \mu_1 - \mu_2 = 0\) vs \(H_0: \mu_1 - \mu_2 \ne 0\)

```
> # compute critical value
> qt(0.05/2, df = 17, ncp = 0, lower.tail = FALSE)
[1] 2.109816
> # perform hypothesis testing
> t.test(x = cFresh, y = cStored, alternative = "two.sided", mu = 0,
+ paired = FALSE, var.equal = FALSE, conf.level = 0.95)
Welch Two Sample t-test
data: cFresh and cStored
t = 4.2368, df = 16.628, p-value = 0.000581
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.2706369 0.8093631
sample estimates:
mean of x mean of y
10.37 9.83
```

At Type I error probability \(\alpha=0.02\)

- Test \(H_0: \mu_1 - \mu_2 \le 0.2\) vs \(H_0: \mu_1 - \mu_2 > 0.2\)

```
> # compute critical value
> qt(0.02, df = 17, ncp = 0, lower.tail = FALSE)
[1] 2.223845
> # perform hypothesis testing
> t.test(x = cFresh, y = cStored, alternative = "greater", mu = 0.2,
+ paired = FALSE, var.equal = FALSE, conf.level = 0.99)
Welch Two Sample t-test
data: cFresh and cStored
t = 2.6676, df = 16.628, p-value = 0.008227
alternative hypothesis: true difference in means is greater than 0.2
99 percent confidence interval:
0.2120818 Inf
sample estimates:
mean of x mean of y
10.37 9.83
```

Are average estimates for repair costs from Garage I different than Garage II?

```
> # estimates of cost from Garage I
> GarageI = c(17.6, 20.2, 19.5, 11.3, 13, 16.3, 15.3, 16.2, 12.2,
+ 14.8, 21.3, 22.1, 16.9, 17.6, 18.4)
>
> # estimates of cost from Garage I
> GarageII = c(17.3, 19.1, 18.4, 11.5, 12.7, 15.8, 14.9, 15.3,
+ 12, 14.2, 21, 21, 16.1, 16.7, 17.5)
```

Boxplot of difference between estimates

Test \(H_0: \mu_1 - \mu_2 = 0\) vs \(H_a: \mu_1 - \mu_2 \ne 0\)

- Try t test based on independent samples
- test statistic value: 0.54616
- degrees of freedom: 27.797
- p-value of test: 0.5893

Conclusion …

Histogram for differences between the estimates

What is wrong with applying t test based on independent samples to this data set?

- Are the two samples independent?
- For each car, are the two estimates for it independent?
- Normality violated?

On observations:

- Sample 1: \(y_{1i}, i=1,\ldots,n\)
- Sample 2: \(y_{2i}, i=1,\ldots,n\)
- Differences: \(d_{i}= y_{1i} - y_{2i}, i=1,\ldots,n\)

Requirements:

- Sampling distribution of \(d_{i}\)’s is Normal
- The \(d_{i}\)’s are independent

- Obtain \(\bar{d}\) and \(s_d\), the sample mean and stardand deviation of \(d_{i}\)’s
\(D_0\): a specified value on \(\mu_d = \mu_1 - \mu_2\)

Test statistic: \(T = \frac{\bar{d} - D_0}{s_d/\sqrt{n}}\) follows a t distribution with \(\textrm{df}=n-1\)

Check Normality on differences

p-value of KS test: 0.8008

- Recall \(\mu_d = \mu_1 - \mu_2\)
Assess \(H_0: \mu_d \le 0\) vs \(H_a: \mu_d >0\)

- Observed T.S. value: \(t = \frac{0.613-0}{0.394/\sqrt{15}} = 6.026\)
- Critical value at Type I error probability \(\alpha = 0.05\) is \(t_{0.05,14}=1.761\); T table
Reject \(H_0\) if \(T \ge t_{\alpha,n-1}\)

The \((1-\alpha)\%\) CI for \(\mu_d = \mu_1 - \mu_2\) is \[\bar{d} \pm t_{\alpha/2,n-1}\frac{s_d}{\sqrt{n}}\]

Computing CI:

- \(\bar{d} = 0.613\), \(s_d = 0.394\), \(t_{0.025,14}=2.14\)
- 95% CI is: \(0.613 \pm 2.14\times \frac{0.394}{\sqrt{15}}\), i.e., \([0.395,0.831]\)

The \((1-\alpha)\%\) CI for \(\mu_d = \mu_1 - \mu_2\) is \[\bar{d} \pm t_{\alpha/2,n-1}\frac{s_d}{\sqrt{n}}\]

Construct 99% CI:

- \(n = 12\), \(\bar{d} = 0.5\), \(s_d = 0.49\),
- \(t_{0.01,14}\) T table

- Test statistic: \(T = \frac{\bar{d} - D_0}{s_d/\sqrt{n}}\) follows a t distribution with \(\textrm{df}=n-1\)
- Recall \(\mu_d = \mu_1 - \mu_2\)
At Type I error prob \(\alpha = 0.05\), test \(H_0: \mu_d= 0.5\) vs \(H_a: \mu_d \ne 0.5\)

- \(n = 16\), \(\bar{d} = 0.7\), \(s_d = 1\),
- \(D_0 =0.5\) and \(t_{0.025,16}\); T table
Reject \(H_0\) when \(|T| > t_{\alpha/2,df}\)

```
> RepairCost = read.table("http://math.wsu.edu/faculty/xchen/stat412/data/pairedT.txt",
+ sep = "\t", header = T)
> RepairCost[1:3, ]
GarageI GarageII
1 17.6 17.3
2 20.2 19.1
3 19.5 18.4
> GarageI = RepairCost$GarageI
> GarageII = RepairCost$GarageII
```

Construct confidence interval

```
> t.test(GarageI, GarageII, alternative = "two.sided", mu = 0,
+ paired = TRUE, var.equal = FALSE, conf.level = 0.95)
Paired t-test
data: GarageI and GarageII
t = 6.0234, df = 14, p-value = 3.126e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.3949412 0.8317254
sample estimates:
mean of the differences
0.6133333
```

Test on difference \(\mu_d = \mu_1 - \mu_2\)

- Test \(H_0: \mu_d \le 0\) vs \(H_a: \mu_d >0\)
- Reject \(H_0\) if \(T \ge t_{\alpha,df}\)

```
> t.test(GarageI, GarageII, alternative = "greater", mu = 0, paired = TRUE,
+ var.equal = FALSE, conf.level = 0.95)
Paired t-test
data: GarageI and GarageII
t = 6.0234, df = 14, p-value = 1.563e-05
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
0.4339886 Inf
sample estimates:
mean of the differences
0.6133333
```

Conclusion?

Test on difference \(\mu_d = \mu_1 - \mu_2\)

- Test \(H_0: \mu_d = 0.4\) vs \(H_a: \mu_d \ne 0.4\)
- Reject \(H_0\) if \(|T| \ge t_{\alpha/2,df}\)

```
> t.test(GarageI, GarageII, alternative = "two.sided", mu = 0.4,
+ paired = TRUE, var.equal = FALSE, conf.level = 0.95)
Paired t-test
data: GarageI and GarageII
t = 2.0951, df = 14, p-value = 0.05483
alternative hypothesis: true difference in means is not equal to 0.4
95 percent confidence interval:
0.3949412 0.8317254
sample estimates:
mean of the differences
0.6133333
```

Conclusion?

Test on difference \(\mu_d = \mu_1 - \mu_2\)

- Test \(H_0: \mu_d \ge 0.3\) vs \(H_a: \mu_d < 0.3\)
- Reject \(H_0\) if \(T \le - t_{\alpha,df}\)

```
> t.test(GarageI, GarageII, alternative = "less", mu = 0.3, paired = TRUE,
+ var.equal = FALSE, conf.level = 0.95)
Paired t-test
data: GarageI and GarageII
t = 3.0772, df = 14, p-value = 0.9959
alternative hypothesis: true difference in means is less than 0.3
95 percent confidence interval:
-Inf 0.7926781
sample estimates:
mean of the differences
0.6133333
```

Conclusion?

- The rejection regions are similar to those for t test based on independent samples
- Each pair of observations are usually obtained from the same individual or item (i.e., from the same experimental unit)
- For each experimental unit, the pair of observations are dependent
- The pairs of observations are independent

- Test is applied to differences obtained from the pairs
- Test is essentially a one-sample t test where sample is the pairwise differences
- Rejection regions are very similar to those of one-sample t test

- Measurements on the same experimental unit obtained before and after receiving a treatement, e.g., in assessing drug effect
- Measurements obtained on two experimental units with similar features, e.g., in assessing improvement of new teaching methods

- Variation of potency for drug
- Risks in portfolios
- Assess equality of variances in two-sample test

Random sample \(y_1,\ldots,y_n\) with mean \(\mu\) and variance \(\sigma^2\)

Sample mean \(\bar{y}=\frac{1}{n}\sum_{i=1}^n y_i\)

Sample variance \(s^2 = \frac{1}{n-1}\sum_{i=1}^n (y_i - \bar{y})^2\)

- Random sample 1: \(y_{1i}, i=1,\ldots,n_1\) follow \(\mathsf{N}(\mu_1, \sigma_1^2)\)
- Random sample 2: \(y_{21}, i=1,\ldots,n_2\) follow \(\mathsf{N}(\mu_2, \sigma_2^2)\)
- Sample variance \(s_1^2\) for random sample 1; sample variance \(s_2^2\) for random sample 2

Then \[\frac{s^2/\sigma_1^2}{s_2^2/\sigma_2^2}\] follows an F distribution

Density: \(df_1=3\) and \(df_2=5\)

- not symmetrical
- with non-negative values
- with \(df_1\) (for \(s_1^2\)) and \(df_2\) (for \(s_2^2\))

Test statistic \(F= \frac{s_1^2}{s_2^2}\)

For \(H_0: \sigma_1^2 \le \sigma_2^2\) vs \(H_a: \sigma_1^2 > \sigma_2^2\), reject \(H_0\) if \(F \ge F_{\alpha,df_1,df_2}\)

For \(H_0: \sigma_1^2= \sigma_2^2\) vs \(H_a: \sigma_1^2 \ne \sigma_2^2\), reject \(H_0\) if \(F \ge F_{\alpha/2,df_1,df_2}\) or \(F \le F_{1-\alpha/2,df_1,df_2}\)

F table; \(F_{1-\alpha,df_1,df_2}= \frac{1}{F_{\alpha,df_2,df_1}}\)

\(100(1-\alpha)\%\) confidence interval for \(\sigma_1^2/\sigma_2^2\) is constructed as follows:

- obtain \(s_1^2\), \(s_2^2\) and \(s_1^2/s_2^2\)
- obtain \(df_1 = n_1 -1\) and \(df_2 = n_2 -1\)
obtain \(F_U = F_{\alpha/2,df_2,df_1}\) and \(F_L = 1/F_{\alpha/2,df_1,df_2}\)

confidence interval: \(\left[\frac{s_1^2}{s_2^2}F_L,\frac{s_1^2}{s_2^2}F_U\right]\)

Data:

```
> # potency readings for Fresh drug
> cFresh = c(10.2, 10.5, 10.3, 10.8, 9.8, 10.6, 10.7, 10.2, 10,
+ 10.6)
>
> # potency readings for Stored drug
> cStored = c(9.8, 9.6, 10.1, 10.2, 10.1, 9.7, 9.5, 9.6, 9.8, 9.9)
```

Nomality test

At Type I error probability \(\alpha = 0.05\), test \(H_0: \sigma_1^2= \sigma_2^2\) vs \(H_a: \sigma_1^2 \ne \sigma_2^2\)

- Sample 1: Fresh; Sample 2: Stored
- \(n_1 = 10\), \(n_2 = 10\); \(s_1^2 = 0.10\), \(s_2^2= 0.06\)
- \(F = \frac{s_1^2}{s_2^2} = 1.67\)
- \(F_{0.025,10,10} = 3.72\); \(F_{0.975,10,10} =1/3.72= 0.27\); F table

Reject \(H_0\) if \(F \ge F_{\alpha/2,df_1,df_2}\) or \(F \le F_{1-\alpha/2,df_1,df_2}\). Conclusion?

\(95\%\) confidence interval for \(\sigma_1^2/\sigma_2^2\)

- \(n_1 = 10\), \(n_2 = 10\)
- \(s_1^2 = 0.10\), \(s_2^2= 0.06\) and \(\frac{s_1^2}{s_2^2} = 1.67\)
- \(F_U = F_{0.025,10,10} = 3.72\); \(F_L = F_{0.975,10,10} = 0.27\)
- CI: \([1.67\times 0.27, 1.67\times 3.72]\)

- If you use software to obtain answers, please attach the codes
- Please put codes close to their associated answers
- Homework assignments will be announced via Blackboard

```
> sessionInfo()
R version 3.3.0 (2016-05-03)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 15063)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] knitr_1.17
loaded via a namespace (and not attached):
[1] backports_1.1.0 magrittr_1.5 rprojroot_1.2 formatR_1.5
[5] tools_3.3.0 htmltools_0.3.6 revealjs_0.9 yaml_2.1.14
[9] Rcpp_0.12.12 stringi_1.1.5 rmarkdown_1.6 stringr_1.2.0
[13] digest_0.6.12 evaluate_0.10.1
```