Answer:
t = -2.69 , p = 0.0078
Step-by-step explanation:
We have the following data:
Sample Mean = x = 24.444
Sample Standard Deviation = s = 2.45
Sample size = n = 18
Josh wants to test that mean number is lesser than 26. So our test value is 26 i.e.
u = 26
Since Josh wants to test that mean would be fewer than 26, so this would be a left tailed test with a less than sign in Alternate Hypothesis. Therefore, the hypothesis would be:
[tex]H_{o}: u\geq 26\\H_{a}: u<26[/tex]
Since, we do not know the value of Population standard deviation, and we have the value of sample standard deviation, we will use One-Sample t-test for the population mean.
The formula to calculate the test-statistic would be:
[tex]t=\frac{x-u}{\frac{s}{\sqrt{n}}}[/tex]
Substituting the values in this formula gives us:
[tex]t=\frac{24.444-26}{\frac{2.45}{\sqrt{18}}}\\t=-2.69[/tex]
This means, the test statistic would be -2.69
Since, the sample size is 18, the degrees of freedom would be:
Degrees of freedom = df = n - 1 = 17
To find the p-value we need to check the p-value against test statistic of 2.69, with 17 degrees of freedom and One-tailed test. This value comes out to be:
p-value = 0.0078
Therefore, the correct answer would be:
t = -2.69 , p = 0.0078
