A simple random sample of size nequals40 is drawn from a population. The sample mean is found to be 103.9β, and the sample standard deviation is found to be 21.9. Is the population mean greater than 100 at the alphaequals0.05 level ofβ significance?
Accepted Solution
A:
Answer:There is enough evidence to support the claim that the population mean is greater than 100Step-by-step explanation:Step 1: We state the hypothesis and identify the claim[tex]H_0:\mu=100[/tex] and [tex]H_1:\mu \:>\:100[/tex] (claim)Step 2: Calculate the test value.[tex]t=\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex][tex]\implies t=\frac{103.9-100}{\frac{21.9}{\sqrt{40} } }=1.1263[/tex]Step 3: Find the P-value. The p-value obtained from a calculator is using d.f=39 and test-value 1.126 is 0.134Step 4: We fail to reject the null hypothesis since P-value Β is greater that the alpha level. (0.134>0.05).Step 5: There is enough evidence to support the claim that the population mean is greater than 100.Alternatively: We could also calculate the critical value to obtain +1.685 for [tex]\alpha=0.05[/tex] and d.f=39 and compare to the test-value:The critical value (1.685>1.126) falls in the non-rejection region. See attachment.NB: The t- distribution must be used because the population standard deviation is not known.