Question

asked Dec 26, 2021 230k views

1 Answer

Tyrone · Answer 1 · 2022-01-01T05:26:08+0000

Answer:

t=(46-40)/((5.148)/(√(5)))=2.606

The degrees of freedom are given by:

df=n-1=5-1=4

The p value wuld be given by:

p_v =2*P(t_((4))>2.606)=0.060

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

$\bar X= (\sum_(i=1)^n X_i)/(n)$

$s =\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}$

$\bar X=46$ represent the mean height for the sample

s=5.148 represent the sample standard deviation

n=5 sample size

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:

Null hypothesis:
$\mu = 40$

Alternative hypothesis:
$\mu \\eq 40$

The statistic is given by:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (1)

Replacing we got: