An educator claims that the average salary of substitute teachers in Pennsylvania is at most $60 per day. A random sample of ten school districts is selected, and the daily salaries are shown. Is there evidence to support the claim at á = .01?

48 56 66 55 70 51 58 49 69 62

Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

I assume you are referring to SPSS in your subject line, which is a computer program for analyzing statistical data. If so, conduct a one-sample t-test using the data you have entered in your post. If you are unfamiliar with SPSS, there are online tutorials which may help.

An educator claims that the average salary of substitute teachers in school districts in Allegheny County, Pennsylvania, is less than $60 a day. A random sample of eight school Districts is selected, and the daily salaries (in dollars) are shown:

60 56 60 55 70 55 60 55
a) Is there enough evidence to support the educator’s claim at α=0.10

Maths biostatics

To determine if there is evidence to support the claim that the average salary of substitute teachers in Pennsylvania is at most $60 per day, we can perform a hypothesis test.

The null hypothesis (H0) would be that the average salary is $60 per day or higher, and the alternative hypothesis (Ha) would be that the average salary is less than $60 per day.

H0: μ ≥ $60 (null hypothesis)
Ha: μ < $60 (alternative hypothesis)

To test this, we can use a one-sample t-test since we have the sample mean and want to compare it to a known or hypothesized population mean.

Here are the steps to perform the hypothesis test:

1. Calculate the sample mean (x̄) of the daily salaries.
x̄ = (48 + 56 + 66 + 55 + 70 + 51 + 58 + 49 + 69 + 62) / 10 = 58.4

2. Calculate the sample standard deviation (s) of the daily salaries.
s = √[(∑(xi - x̄)^2) / (n - 1)] = √[(∑(xi - 58.4)^2) / 9]

3. Define the level of significance (α) as 0.01.
α = 0.01

4. Determine the critical t-value for a one-tailed test with 9 degrees of freedom (n - 1) and α = 0.01. You can look up the critical value from a t-distribution table or use statistical software.

5. Calculate the t-value using the formula:
t = (x̄ - μ) / (s / √n)
where μ is the hypothesized population mean ($60) and n is the sample size (10).

6. Compare the calculated t-value with the critical t-value.

- If the calculated t-value is less than the critical t-value, we reject the null hypothesis and conclude that there is evidence to support the claim.
- If the calculated t-value is greater than or equal to the critical t-value, we fail to reject the null hypothesis and do not have sufficient evidence to support the claim.

Performing the calculations, we find:
t = (58.4 - 60) / (s / √10)

To complete the calculations, we need the value of s from step 2. You can calculate it using the given sample values.

After calculating the t-value and comparing it to the critical t-value, you can make a conclusion about whether there is evidence to support the claim.