![PSet5 Question 7
TA Kunyu (Quinn) He CAPP’20
2018.12.06
a = c(1.1650, 0.6268, 0.0751, 0.3516)
b = c(0.3035, 2.6861, 1.0591, 2.7971, 1.2641)
(a)
mu_a = mean(a)
mu_b = mean(b)
diff = mu_a - mu_b
diff
## [1] -1.067355
(b)
According to the pooled sample variance formula:
S_p = ((length(a)-1) * var(a) + (length(b)-1) * var(b)) / (length(a) + length(b) - 2)
S_p
## [1] 0.7636414
(c)
se = sqrt(S_p/length(a) + S_p/length(b))
se
## [1] 0.586207
(d)
Critical value at α = 0.05 and d.f. = 7:
qt(0.95, 7)
## [1] 1.894579
CI:
c(diff - qt(0.95, 7)*se, diff + qt(0.95, 7)*se)
## [1] -2.17797025 0.04326025
(e)
Here I test both positive and negative difference, since you might make your hypotheses in
different ways. But both ways come to the same result.
2 * pt(diff / se, df = 7)
## [1] 0.1114408
p-value is the probability, assuming that null hypothesis is correct, that we observe more extreme values than
our test statistic. In this case our test statistic is obtained by difference in the mean divided by standard
error as below:
diff / se
1](https://image.slidesharecdn.com/pset5question7-181206173006/85/P-set5-question_7-1-320.jpg)
![## [1] -1.820782
So here the test statistic is negative, pt function with df (degree of freedom) equals to 7 is giving back the
area under the pdf curve, left to -1.820782. This denotes the probability, assuming 0 difference, we observe
values lower than -1.820782.
We multiply this by 2 to get the p-value, as we are doing two-tailed hypothesis testing we need to take
probability of observing larger values than 1.820782 into consideration. Since pdf is symmetric, the probability
of observing larger values is the same.
2 * pt(-diff / se, df = 7, lower.tail = F)
## [1] 0.1114408
Since some of you go with positive difference. Here I tested it, assigning lower.tail=F, which gives me the
area under the pdf curve, right to the test statistic(1.820782). Multiply it by 2, we get the same p-value.
When it comes to finding the probability with t-table, if you’re testing a negative test-statistic, go to the t-table
with its absolute value, get the corresponding probability. Since t-table always give back upper-tail
probability, what you need to do is just multiply that by 2.
Always useful to have the pdf in mind:
curve(dt(x, 7), from = -4, to = 4, col = "orange",
xlab = "Quantile", ylab = "Density", lwd = 2)
−4 −2 0 2 4
0.00.10.20.30.4
Quantile
Density
(h)
2](https://image.slidesharecdn.com/pset5question7-181206173006/85/P-set5-question_7-2-320.jpg)
![S = 1
se = sqrt(S/length(a) + S/length(b))
se
## [1] 0.6708204
CI = c(diff - qnorm(0.95)*se, diff + qnorm(0.95)*se)
CI
## [1] -2.17075636 0.03604636
p-value:
2 * pnorm(diff / se)
## [1] 0.1115828
pdf of standard normal distribution:
curve(dnorm(x), from = -4, to = 4, col = "blue",
xlab = "Quantile", ylab = "Density", lwd = 2)
−4 −2 0 2 4
0.00.10.20.30.4
Quantile
Density
3](https://image.slidesharecdn.com/pset5question7-181206173006/85/P-set5-question_7-3-320.jpg)
P set5 question_7
- 1. PSet5 Question 7 TA Kunyu (Quinn) He CAPP’20 2018.12.06 a = c(1.1650, 0.6268, 0.0751, 0.3516) b = c(0.3035, 2.6861, 1.0591, 2.7971, 1.2641) (a) mu_a = mean(a) mu_b = mean(b) diff = mu_a - mu_b diff ## [1] -1.067355 (b) According to the pooled sample variance formula: S_p = ((length(a)-1) * var(a) + (length(b)-1) * var(b)) / (length(a) + length(b) - 2) S_p ## [1] 0.7636414 (c) se = sqrt(S_p/length(a) + S_p/length(b)) se ## [1] 0.586207 (d) Critical value at α = 0.05 and d.f. = 7: qt(0.95, 7) ## [1] 1.894579 CI: c(diff - qt(0.95, 7)*se, diff + qt(0.95, 7)*se) ## [1] -2.17797025 0.04326025 (e) Here I test both positive and negative difference, since you might make your hypotheses in different ways. But both ways come to the same result. 2 * pt(diff / se, df = 7) ## [1] 0.1114408 p-value is the probability, assuming that null hypothesis is correct, that we observe more extreme values than our test statistic. In this case our test statistic is obtained by difference in the mean divided by standard error as below: diff / se 1
- 2. ## [1] -1.820782 So here the test statistic is negative, pt function with df (degree of freedom) equals to 7 is giving back the area under the pdf curve, left to -1.820782. This denotes the probability, assuming 0 difference, we observe values lower than -1.820782. We multiply this by 2 to get the p-value, as we are doing two-tailed hypothesis testing we need to take probability of observing larger values than 1.820782 into consideration. Since pdf is symmetric, the probability of observing larger values is the same. 2 * pt(-diff / se, df = 7, lower.tail = F) ## [1] 0.1114408 Since some of you go with positive difference. Here I tested it, assigning lower.tail=F, which gives me the area under the pdf curve, right to the test statistic(1.820782). Multiply it by 2, we get the same p-value. When it comes to finding the probability with t-table, if you’re testing a negative test-statistic, go to the t-table with its absolute value, get the corresponding probability. Since t-table always give back upper-tail probability, what you need to do is just multiply that by 2. Always useful to have the pdf in mind: curve(dt(x, 7), from = -4, to = 4, col = "orange", xlab = "Quantile", ylab = "Density", lwd = 2) −4 −2 0 2 4 0.00.10.20.30.4 Quantile Density (h) 2
- 3. S = 1 se = sqrt(S/length(a) + S/length(b)) se ## [1] 0.6708204 CI = c(diff - qnorm(0.95)*se, diff + qnorm(0.95)*se) CI ## [1] -2.17075636 0.03604636 p-value: 2 * pnorm(diff / se) ## [1] 0.1115828 pdf of standard normal distribution: curve(dnorm(x), from = -4, to = 4, col = "blue", xlab = "Quantile", ylab = "Density", lwd = 2) −4 −2 0 2 4 0.00.10.20.30.4 Quantile Density 3