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 0.01033Reference
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].