Question

Question 1 options:Residents in Portland, Oregon think that their city has more rainfall than Seattle, Washington. To test this claim, citizens collect data on annual rainfall. In Portland, it is found that the average rainfall over 45 years is 37.50 inches, with a standard deviation of 1.82 inches. In Seattle, the average annual rainfall over 35 years is 37.07 inches, with a standard deviation of 1.68 inches. Is there enough evidence to support the claim that Portland has more average yearly rainfall than Seattle using a level of significance of 10%?Enter the Null Hypothesis for this test: H0:Enter the Alternative Hypothesis for this test: H1:What is the p-value for this hypothesis test? Round your answer to four decimal places.What is the decision based on the given sample statistics?

Answer 1

Answer:

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0$

P-value = 0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

Step-by-step explanation:

We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0$

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

The difference between means is:

$M_d= M_1-M_2=37.50-37.07=0.43$

The standard error for the difference between means is:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774$

We can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393$

The degrees of freedom are:

$df=n1+n2-2=45+35-2=78$

Then, the p-value for this one-tailed test with 78 degrees of freedom is:

$P-value=P(t>1.1393)=0.1290$

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.