Question

An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample of 8 school districts is selected, and the daily salaries are 60, 56, 60, 55, 70, 55, 60, and 55. Is there enough evidence to support the educator’s claim at 10% level of significance? (HELP: The sample mean is 58.88, and the sample standard deviation is 5.08)

Answer 1

`t=(58.875-60)/((5.083)/(√(8)))=-0.626`

The degrees of freedom are given by:

`df=n-1=8-1=7`

The p value would be given by:

`p_v =P(t_((7))<-0.626)=0.275`

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation 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)}`

Replacing we got:

`\bar X=58.875` represent the mean

`s=5.083` represent the sample standard deviation for the sample

`n=8` sample size

`\mu_o =60` represent the value that we want to test

`\alpha=0.1` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 60`

Alternative hypothesis:
`\mu < 60`

The statistic would be given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(58.875-60)/((5.083)/(√(8)))=-0.626`

The degrees of freedom are given by:

`df=n-1=8-1=7`

The p value would be given by:

`p_v =P(t_((7))<-0.626)=0.275`

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60