Question

Use the null hypothesis H0 : μ = 98.6, alternative hypothesis Ha: μ < 98.6, and level of significance α = 0.05. 98 99.6 97.8 97.6 98.7 98.4 98.9 97.1 99.2 97.4 99.1 96.9 98.8 99.9 96.8 97 98.7 97.6 98.7 98.2 whats the t-score?

Answer 1

Answer:

The t-score is -1.8432    

Step-by-step explanation:

We are given the following in the question:  

98, 99.6, 97.8, 97.6, 98.7, 98.4, 98.9, 97.1, 99.2, 97.4, 99.1, 96.9, 98.8, 99.9, 96.8, 97, 98.7, 97.6, 98.7, 98.2

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1964.4}{20} = 98.22$

Sum of squares of differences = 16.152

$s = \sqrt{\dfrac{16.152}{49}} = 0.922$

Population mean, μ = 98.6

Sample mean, $\bar{x}$ = 98.22

Sample size, n = 20

Sample standard deviation, s = 0.922

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 98.6\\H_A: \mu < 98.6$

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{98.22 - 98.6}{\frac{0.922}{\sqrt{20}} } = -1.8432$

The t-score is -1.8432