# Hypothesis test for a difference between means

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## Given Problem : One-Tailed Test

## Hypothesis test

This lesson explains how to conduct a hypothesis test for the difference between two means. The test procedure, called the two-sample t-test, is appropriate when the following conditions are met:

- The sampling method for each sample is simple random sampling.

- The samples are independent.

- Each population is at least 20 times larger than its respective sample.

- The sampling distribution is approximately normal, which is generally the case if any of the following conditions apply.

  - The population distribution is normal.

  - The population data are symmetric, unimodal, without outliers, and the sample size is 15 or less.

  - The population data are slightly skewed, unimodal, without outliers, and the sample size is 16 to 40.

  - The sample size is greater than 40, without outliers.

### This approach consists of four steps:

1. state the hypotheses

2. formulate an analysis plan

3. analyze sample data

4. interpret results.

### 1. State the hypotheses

The first step is to state the null hypothesis and an alternative hypothesis.

\[Null\ hypothesis(H_0): \mu_1-\mu_2 <= 7\] \[Alternative\ hypothesis(H_1): \mu_1-\mu_2 > 7\]

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the difference between sample means is too big..

### 2. Formulate an analysis plan.

For this analysis, the significance level is 95%. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

### 3. Analyze sample data

Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

\[SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}\] \[SE=\sqrt{\frac{40^2}{100}+\frac{20^2}{100}}\] \[SE=4.472\]

\[DF=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}\]

\[DF=\frac{(\frac{40^2}{100}+\frac{20^2}{100})^2}{\frac{(40^2/100)^2}{100-1}+\frac{(20^2/100)^2}{100-1}}\] \[DF=145.59\]

\[t=\frac{(\bar{x_1}-\bar{x_2})-d}{SE} = \frac{(200 -190)-7}{4.472}=0.671\]

where \(s_1\) is the standard deviation of sample 1, \(s_2\) is the standard deviation of sample 2, \(n_1\) is the size of sample 1, \(n_2\) is the size of sample 2, \(\bar{x_1}\) is the mean of sample 1, \(\bar{x_2}\) is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

We can plot the mean difference.

    plot(x)

Since we have a one-tailed test, the P-value is the probability that the t statistic having 145.59 degrees of freedom is or greater than 0.67.

We use the t Distribution curve to find p value.

    draw_t(DF=round(x$result$DF,2),t=x$result$t,alternative=x$result$alternative)

### 4. Interpret results.

Since the P-value (0.252) is greater than the significance level (0.05), we cannot reject the null hypothesis.

### Result of meanCI()

    call: meanCI.default(n1 = 100, n2 = 100, m1 = 200, s1 = 40, m2 = 190,      s2 = 20, mu = 7, alpha = 0.05, alternative = "greater")
    method: Welch Two Sample t-test
    alternative hypothesis:
       true unpaired differences in means is greater than  7

    Results
    # A tibble: 1 × 6
      control test  DF     CI                     t       p
      <chr>   <chr> <chr>  <chr>                  <chr>   <chr>
    1 x       y     145.59 10.00 [95CI 2.60; Inf] 0.67082 0.2517

### Reference

The contents of this document are modified from StatTrek.com. Berman H.B., “AP Statistics Tutorial”, [online] Available at: https://stattrek.com/hypothesis-test/difference-in-means.aspx?tutorial=AP URL[Accessed Data: 1/23/2022].