Question

Consider the hypothesis test given by

Answer 1

Answer:

t = -1.633<2.06

Step 1

Here χ is the sample mean

μ  is the population mean

S  is the sample standard deviation

n be the sample size

The degrees of freedom γ =n-1

Step 2:-

a) we will use t- distribution test

Here sample size n = 25

the sample mean χ =640

population mean μ =650

sample standard deviation S=30

Step 3:-

b) we will use two tailed test or left tailed test

Null Hypothesis H0 : μ=650

Alternative Hypothesis Ha:μ<650

$t = \frac{x-u}{\frac{S}{\sqrt{n-1} } }$

Step 4:-

c)

$t = \frac{640-650}{\frac{30}{\sqrt{25-1} } }$

[tex]t = -1.633

Step 5 :-

Degrees of freedom:-

The degrees of freedom of t- distribution

γ = n-1 = 25-1 = 24

The table value of t are 5% level with 24 degrees of freedom For two tailed test is 2.06

Step 6:-

Since the calculated value of t < tabulated value of t, so we accepted the null hypothesis.

The data support the assumption of a population is normally distributed