This lesson is about testing hypotheses about the difference between two population means, \(\mu_1\) and \(\mu_2\). We will consider three different cases depending on whether the samples are independent or dependent and also depending on whether the population variances are equal.
When the samples are dependent, we analyze the differences. More on this below in Case 3.
When the samples are independent, we use the sampling distribution of \(\bar X_1 - \bar X_2\) as the basis of the confidence interval.
\[\bar X_1 - \bar X_2 \sim N(\mu_1 - \mu_2,\sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} })\] Of course, we donโt know \(\sigma_1^2\) and \(\sigma_2^2\) so we are going to use the sample variances \(s_1^2\) and \(s_2^2\) which means that we will be using the \(t\) distribution instead of the normal distribution.
The degrees of freedom and the exact form of the estimated standard error depends on whether we have equal population variances.
Step 1: Check conditions required
Step 2: Define the population means, \(\mu_1\) and \(\mu_2\), in the context of the problem.
Step 3: Determine the null and alternative hypotheses. They can be structured three ways.
| Two tailed | Left tailed | Right tailed |
|---|---|---|
| \(H_0: \mu_1 = \mu_2\) | \(H_0: \mu_1 = \mu_2\) | \(H_0: \mu_1 = \mu_2\) |
| \(H_a: \mu_1 \ne \mu_2\) | \(H_a: \mu_1 < \mu_2\) | \(H_0: \mu_1 > \mu_2\) |
Step 4: Compute the test statistic
\[t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\]
Step 5: Find the p-value with approximate degrees of freedom = min\((n_1-1,n_2-1)\). Software such as R will use exact degrees of freedom which is a more complex formula.
| Two tailed | Left tailed | Right tailed |
|---|---|---|
| \(2P(T>|t|)\) = 2*(1-pt(abs(t),df)) | \(P(T<t)\) = pt(t,df) | \(P(T>t)\) = 1 - pt(t,df) |
Step 6: Reject \(H_0\) if p-value\(\le \alpha\). Otherwise, fail to reject \(H_0\)
Step 7: State your conclusions in the context of the problem.
If you have raw data, then you can use the t.test command to find the test statistic and p-value.
Example 1:
The local baseball team conducts a study to find the amount spent on refreshments at the ball park. Over the course of the season, they gather simple random samples of 20 men and 23 women. For men, the average expenditure was $20 with a standard deviation of $3. For women, the average expenditure was $18 with a standard deviation of $2. Do men and women spend differently on average? Test at \(\alpha = 0.01.\)
Click For AnswerStep 1: We require the following conditions
Step 2: Let \(\mu_1=\) mean expenditure for men and \(\mu_2=\) mean expenditure for women.
Step 3: We want to test the null hypothesis of no difference in spending on average \(H_0: \mu_1 = \mu_2\) against the alternative hypothesis that there is a difference in spending on average \(H_a: \mu_1 \ne \mu_2\)
Step 4: To find the test statistic we first enter the sample data
n1 <- 20
xbar1 <- 20
s1 <- 3
n2 <- 23
xbar2 <- 18
s2 <- 2Compute the standard error
se <- sqrt(s1^2/n1 + s2^2/n2)
se[1] 0.7898817Finally, find the test statistic
t <- (xbar1 - xbar2)/se
t[1] 2.532025Step 5: To find the p-value, we need to look at the direction of the test. In this case, we are conducting a two tailed test.
df <- min(n1-1,n2-1)
2*(1 - pt(abs(t),df))[1] 0.02031837Step 6: Fail to reject \(H_0\) since the p-value of \(0.02\) > \(\alpha = 0.01\).
Step 7: We have insufficient evidence to conclude that there is a significant difference in mean expenditure for men and women at 1% significance.
Step 1: Check conditions required
Step 2: Define the population means, \(\mu_1\) and \(\mu_2\), in the context of the problem.
Step 3: Determine the null and alternative hypotheses. They can be structured three ways.
| Two tailed | Left tailed | Right tailed |
|---|---|---|
| \(H_0: \mu_1 = \mu_2\) | \(H_0: \mu_1 = \mu_2\) | \(H_0: \mu_1 = \mu_2\) |
| \(H_a: \mu_1 \ne \mu_2\) | \(H_a: \mu_1 < \mu_2\) | \(H_0: \mu_1 > \mu_2\) |
Step 4: Compute the test statistic
\[t = \frac{\bar x_1 - \bar x_2}{\sqrt{s_p^2({\frac{1}{n_1}+\frac{1}{n_2}})}}\] where \[s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}\]
Step 5: Find the p-value with df \(= n_1 + n_2 - 2\)
| Two tailed | Left tailed | Right tailed |
|---|---|---|
| \(2P(T>|t|)\) = 2*(1-pt(abs(t),df)) | \(P(T<t)\) = pt(t,df) | \(P(T>t)\) = 1 - pt(t,df) |
Step 6: Reject \(H_0\) if p-value\(\le \alpha\). Otherwise, fail to reject \(H_0\)
Step 7: State your conclusions in the context of the problem.
If you have raw data, then you can use the t.test command to find the test statistic and p-value.
Example 2: Is computer processor A faster than computer processor B? Test the appropriate hypotheses at \(\alpha=0.05\). Assume equal population variances.
| Processor | \(n\) | \(\bar x\) | \(s\) |
|---|---|---|---|
| A | \(17\) | \(3004\) | \(74\) |
| B | \(14\) | \(2538\) | \(56\) |
Step 1: We require the following conditions
Step 2: Let \(\mu_1=\) mean speed for Processor A and \(\mu_2=\) mean speed for Processor B
Step 3: We want to test the null hypothesis of no difference in speed on average \(H_0: \mu_1 = \mu_2\) against the alternative hypothesis that Processor A is faster (has greater speed) than Processor B on average \(H_a: \mu_1 > \mu_2\)
Step 4: To find the test statistic we first enter the sample data
n1 <- 17
xbar1 <- 3004
s1 <- 74
n2 <- 14
xbar2 <- 2538
s2 <- 56Then compute the pooled variance and use it to find the standard eror
pool <- ((n1-1)*s1^2 + (n2-1)*s2^2)/(n1+n2-2)
pool[1] 4427.034se <- sqrt(pool*(1/n1 + 1/n2))
se[1] 24.01313Finally, find the test statistic
t <- (xbar1-xbar2)/se
t[1] 19.40605Step 5: To find the p-value, we need to look at the direction of the test. In this case, we are conducting a right tailed test.
df <- n1 + n2 - 2
1 - pt(t,df)[1] 0Step 6: Reject \(H_0\) since the p-value\(=0\) is less than \(\alpha = 0.05\).
Step 7: We have sufficient evidence to conclude that Processor A is faster than Processor B, on average, at 5% significance.
Step 1: Check conditions required
Step 2: Define the population mean difference \(\mu_d=\mu_1-\mu_2\), in the context of the problem.
Step 3: Determine the null and alternative hypotheses. They can be structured three ways.
| Two tailed | Left tailed | Right tailed |
|---|---|---|
| \(H_0: \mu_d = 0\) | \(H_0: \mu_d = 0\) | \(H_0: \mu_d = 0\) |
| \(H_a: \mu_d \ne 0\) | \(H_a: \mu_d < 0\) | \(H_0: \mu_d > 0\) |
Step 4: Compute the test statistic
\[t=\frac{\bar x_d}{s_d/\sqrt{n_d}}\] where \(n_d=\) the number of differences, \(\bar x_d=\) the sample mean of the differences and \(s_d=\) the sample standard deviation of the differences.
Step 5: Find the p-value with df \(= n_d - 1\)
| Two tailed | Left tailed | Right tailed |
|---|---|---|
| \(2P(T>|t|)\) = 2*(1-pt(abs(t),df)) | \(P(T<t)\) = pt(t,df) | \(P(T>t)\) = 1 - pt(t,df) |
Step 6: Reject \(H_0\) if p-value\(\le \alpha\). Otherwise, fail to reject \(H_0\)
Step 7: State your conclusions in the context of the problem.
Example 3: Six people sign up for a weight loss program. Is it effective? Test the appropriate hypothesis at \(\alpha = 0.05\).
| Person | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Before | 136 | 205 | 257 | 238 | 175 | 166 |
| After | 125 | 195 | 250 | 240 | 165 | 160 |
Step 1: Check conditions required
Step 2: \(\mu_d=\) the mean difference (after-before) in weight loss
Step 3: We want to test the null hypothesis that the diet is not effective \(H_0:\mu_d=0\) against the alternative that the diet is effective \(H_a: \mu_d < 0\).
Step 4: To find the test statistic we first enter the sample data
before <- c(136,205,257,238,175,166)
after <- c(125,195,250,240,165,160)We can use the t.test command with the paired=TRUE subcommand
t.test(after,before,paired=TRUE,alternative="less")
Paired t-test
data: after and before
t = -3.5598, df = 5, p-value = 0.008109
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf -3.037641
sample estimates:
mean of the differences
-7 Or we can find the differences and then find the mean and standard deviation of the differences
dif <- after - before
nd <- length(dif)
xbard <- mean(dif)
sd <- sd(dif)
nd[1] 6xbard[1] -7sd[1] 4.816638Find the test statistic
t <- xbard/(sd/sqrt(nd))
t[1] -3.559833Find the p-value
pt(t,nd-1)[1] 0.008108714Either way we reject the null hypothesis since the p-value of \(0.008\) is less than \(\alpha.\) There is sufficient evidence to conclude that the diet is effective on average.