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------------------------------------------------ A sample of 40 observations is selected from one population with a population standard deviation of 4.6. The sample mean is 101.0. A sample of 47 observations is selected from a second population with a population standard deviation of 4.0. The sample mean is 99.3. Conduct the following test of hypothesis using the 0.10 significance level.

**Hypotheses:**
- \(H_0: \mu_1 = \mu_2\)
- \(H_1: \mu_1 \neq \mu_2\)

**Tasks:**
a. What is the decision rule?

b. Compute the value of the test statistic.

c. What is the p-value?

Answer :

Final answer:

The decision rule for the hypothesis test at a 0.10 significance level involves rejecting the null hypothesis if the test statistic is greater than the critical value. The calculated Z-value is approximately 1.8233, and the corresponding p-value will determine our decision to reject or not reject the null hypothesis.

Explanation:

Decision Rule and Test Statistic

For a hypothesis test comparing two population means with known standard deviations, we use a Z-test for two samples. The decision rule at a significance level of 0.10 involves rejecting the null hypothesis (H0: μ1 = μ2) if the absolute value of the test statistic exceeds the critical value from the standard normal distribution.

Calculation of Test Statistic

The test statistic (Z) is calculated using the formula: Z = (101.0 - 99.3)/ √[ (4.6²/40) + (4.0²/47) ]

Plugging in the numbers: Z = (1.7)/ √[ (21.16/40) + (16/47) ] Z = (1.7)/ √[ 0.529 + 0.3404255 ] Z = (1.7)/√[0.8694255] Z = (1.7)/0.932447 Z ≈ 1.8233

Decision Based on P-value

The p-value associated with this Z-value can be found using standard normal distribution tables or software. If the p-value is less than 0.10, we reject the null hypothesis. Otherwise, we fail to reject it.

Answer:

Step-by-step explanation: