Respuesta :
Answer:
a)
[tex]|z| < 2.054[/tex]: Do not reject the null hypothesis.
[tex]|z| > 2.054[/tex]: Reject the null hypothesis.
b) [tex]z = 2.81[/tex]
c) Reject.
d) The p-value is 0.005.
Step-by-step explanation:
Before testing the hypothesis, we need to understand the central limit theorem and the subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Population 1:
Sample of 42, standard deviation of 3.3, mean of 101, so:
[tex]\mu_1 = 101[/tex]
[tex]s_1 = \frac{3.3}{\sqrt{42}} = 0.51[/tex]
Population 2:
Sample of 53, standard deviation of 3.6, mean of 99, so:
[tex]\mu_2 = 99[/tex]
[tex]s_2 = \frac{3.6}{\sqrt{53}} = 0.495[/tex]
H0 : μ1 = μ2
Can also be written as:
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
H1 : μ1 ≠ μ2
Can also be written as:
[tex]H_1: \mu_1 - \mu_2 \neq 0[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{s}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, and s is the standard error .
a. State the decision rule.
0.04 significance level.
Two-tailed test(test if the means are different), so between the 0 + (4/2) = 2nd and the 100 - (4/2) = 98th percentile of the z-distribution, and looking at the z-table, we get that:
[tex]|z| < 2.054[/tex]: Do not reject the null hypothesis.
[tex]|z| > 2.054[/tex]: Reject the null hypothesis.
b. Compute the value of the test statistic.
0 is tested at the null hypothesis:
This means that [tex]\mu = 0[/tex]
From the samples:
[tex]X = \mu_1 - \mu_2 = 101 - 99 = 2[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.51^2 + 0.495^2} = 0.71[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{s}[/tex]
[tex]z = \frac{2 - 0}{0.71}[/tex]
[tex]z = 2.81[/tex]
c. What is your decision regarding H0?
[tex]|z| = 2.81 > 2.054[/tex], which means that the decision is to reject the null hypothesis.
d. What is the p-value?
Probability that the means differ by at least 2, either plus or minus, which is P(|z| > 2.81), which is 2 multiplied by the p-value of z = -2.81.
Looking at the z-table, z = -2.81 has a p-value of 0.0025.
2*0.0025 = 0.005
The p-value is 0.005.