The right rectangular prisms are similar.

A consumer magazine article titled "How Safe Is the Air in Airplanes" reports that the air quality, as quantified by the degree

OlgaM077 [116]

Answer:

Step-by-step explanation:

Given that a consumer magazine article titled "How Safe Is the Air in Airplanes" reports that the air quality, as quantified by the degree of staleness, was measured on 175 domestic flights.

The purpose was to study about the safety of all flights as a whole.

Since all flights cannot be measured within a reasonable time, a sample of 175 domestic flights were taken.

a) The population of interest here would be the measure of safety of all the airplanes.

b) Sample is the 175 domestic flights selected

c) Characteristics of interest is the safety in travelling by air in Airplanes due to air quality.

8 0

4 years ago

Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last

lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

$t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55$

The degrees of freedom are df=599

$df=n-1=600-1=599$

The P-value for this test statistic is:

$P-value=P(t$

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0

4 years ago

P = -2p - 15<br><br> I need help plz. Thanks!

mihalych1998 [28]

Answer: p=−5

Step-by-step explanation:

Step 1: Add 2p to both sides.

p+2p=−2p−15+2p

3p=−15

Step 2: Divide both sides by 3.

3p/3 = −15/3