Question

Hypothesis Testing: 1. You are given the following information and then asked to do a hypothesis test with a 95% level of confidence to evaluate whether or not it is likely that the population mean is equal to 76 for a given variable. N=10,000 X-bar=74 s2=10,000 2. You are given the following information and then asked to do a hypothesis test with a 95% level of confidence to evaluate whether or not it is likely that the population mean is equal to 440 for a given variable. N=1500 X-bar=450 s2=10,000 3. Explain the differences in these Cls, and why they occur.
please help me answer questions 1 to 3. please show you work clearly with the equations step by step. Thank you.

Answer 1

We would be looking at all the steps of the first question here on hypothesis test.

Q1) As we are testing here whether the population mean is equal to 76, therefore the null and the alternative hypothesis here are given as:

$H_0: \mu =76$

$H_a: \mu \neq 76$

The test statistic here is computed as:

$t^* = \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{74 - 76}{\frac{100}{\sqrt{10000}}} = -2$

as this is a two tailed test, for n - 1 = 9999 degrees of freedom, the p-value here is obtained from t distribution tables as:
p = 2P( t₉₉₉₉ < -2) = 2*0.0228 = 0.0456

As we are doing the test at 95% confidence level, therefore level of significance = 1 - c = 5%.

As the p-value here is 0.0456 < 0.05, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the mean is not equal to 76 here.