L1 : (1, 14) and (1, 2) L2: (5, -2) and (4, 9) A. Parallel B. Neither C. Perpendicular

The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 20

Yuri [45]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=1503\\\\H_a:\mu< 1503$

b) Point estimate d = -$78

c) Test statistic t = -2.438

P-value = 0.0113

Reject H0. We can conclude that the population mean automobile premium in Pennsylvania is lower than the national mean.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that automobile insurance in Pennsylvania is significantly cheaper than the national average.

Then, the null and alternative hypothesis are:

$H_0: \mu=1503\\\\H_a:\mu< 1503$

The significance level is 0.05.

The sample has a size n=25.

The sample mean is M=1425.

A point estimate of the difference between the mean annual premium in Pennsylvania and the national mean can be calculated with the sample mean:

$d=M-\mu=1425-1503=-78$

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=160.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{160}{\sqrt{25}}=32$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1425-1503}{32}=\dfrac{-78}{32}=-2.438$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

This test is a left-tailed test, with 24 degrees of freedom and t=-2.438, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0113) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that automobile insurance in Pennsylvania is significantly cheaper than the national average.

8 0

3 years ago

A least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 46.8 and

Harman [31]

Answer: 31%

Step-by-step explanation:

Formula : Percent of the variability = $R^2\times100=\dfrac{SSR}{SST}\times100$

, where $R^2$ = Coefficient of Determination.

SSR = sum of squares of regression

SST = total sum of squares

$R^2$ is the proportion of the variation of Y that can be attributed to the variation of x.

As per given , we have

SSR = 14.55

SST= 46.8

Then, the percent of the variability in y that can be explained by variability in the regression model = $\dfrac{14.55}{46.8}\times100=31.0897435897\%\approx31\%$