# Hypothesis test for a mean

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## Given Problem : Two-Tailed Test

## Hypothesis Test for a Mean

This lesson explains how to conduct a hypothesis test of a mean, when the following conditions are met:

- The sampling method is simple random sampling.

- The sampling distribution is normal or nearly normal.

Generally, the sampling distribution will be approximately normally distributed if any of the following conditions apply.

- The population distribution is normal.

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

- The population distribution is moderately skewed, unimodal, without outliers, and the sample size is between 16 and 40.

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

## This approach consists of four steps:

- state the hypotheses

- formulate an analysis plan

- analyze sample data

- interpret results.

### 1. State the hypotheses

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

\[Null\ hypothesis(H_0): \mu = 23\] \[Alternative\ hypothesis(H_1): \mu \neq 23\]

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small..

### 2. Formulate an analysis plan

For this analysis, the significance level is 95%. The test method is a one-sample t-test.

### 3. Analyze sample data.

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

\[SE = \frac{s}{\sqrt{n}} = \frac{6.0269481}{\sqrt{32}} = 1.07\] \[DF=n-1=32-1=31\]

\[t = (\bar{x} - \mu) / SE = (20.090625 - 23)/1.07 = -2.731\]

where s is the standard deviation of the sample, \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, and n is the sample size.

We can visualize the confidence interval of mean.

    plot(x)

Since we have a two-tailed test, the P-value is the probability that the t statistic having 31 degrees of freedom is less than -2.73 or greater than 2.73.

We use the t Distribution curve to find p value.

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

\[pt(-2.731,31) =0.01 \]

### 4. Interpret results.

Since the P-value (0.01) is less than the significance level (0.05), we can reject the null hypothesis.

### Result of meanCI()

    call: meanCI.data.frame(x = mtcars, mpg, mu = 23)
    method: One sample t-test
    alternative hypothesis:
       true mean  is not equal to  23

    Results
    # A tibble: 1 × 7
      m        se     DF    lower    upper    t       p
      <chr>    <chr>  <chr> <chr>    <chr>    <chr>   <chr>
    1 20.09062 1.0654 31    17.91768 22.26357 -2.7307 0.01033

### 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/mean.aspx?tutorial=AP URL[Accessed Data: 1/23/2022].