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Phantasy [73]
3 years ago
9

A simple random sample of size nequals40 is drawn from a population. The sample mean is found to be 103.9​, and the sample stand

ard deviation is found to be 21.9. Is the population mean greater than 100 at the alphaequals0.05 level of​ significance?

Mathematics
1 answer:
yan [13]3 years ago
5 0

Answer:

There is enough evidence to support the claim that the population mean is greater than 100

Step-by-step explanation:

<u>Step 1</u>: We state the hypothesis and identify the claim

H_0:\mu=100 and H_1:\mu \:>\:100 (claim)

<u>Step 2</u>: Calculate the test value.

t=\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }

\implies t=\frac{103.9-100}{\frac{21.9}{\sqrt{40} } }=1.1263

<u>Step 3</u>: Find the P-value. The p-value obtained from a calculator is using d.f=39 and test-value 1.126 is 0.134

<u>Step 4</u>: We fail to reject the null hypothesis since P-value  is greater that the alpha level. (0.134>0.05).

<u>Step 5</u>: There is enough evidence to support the claim that the population mean is greater than 100.

<u>Alternatively</u>: We could also calculate the critical value to obtain +1.685 for \alpha=0.05 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.

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