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A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0

3 years ago

A rectangle is plotted on a coordinate plane. The four vertices are located at (−5,5), (−5,−3), (2,5), and (2,−3). What is the a

n200080 [17]

Answer:

56 square units

Step-by-step explanation:

I'd recomend using something like "Desmos" to plot the points so then you can see the points and the rectangle. Then count the length and width. For the this rectangle the length was 8 and the width was 7, multiply these and you get the answer of 56 sqaure units.

8 0

2 years ago

If y represents a student's age, which inequality shows that you must be older than 14 to try out for the basketball team?

Lorico [155]

Answer:

week, but he must work less than ... 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 ... The graph shows the numbers of two types of custom wood tables that can ... Which system of linear inequalities. Custom Tables represents the graph? x + y = 15 ... u=50-2x. 4 = 20. Amount. Ones like not be on test lo+ll. 50-2x320 weeks. 11.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Which of the following is the quotient of b and a?

vlabodo [156]

$\bf 19^{\frac{7}{4}}\cdot \sqrt[a]{19^b}=19^{\frac{5}{2}}\sqrt{19}\\\\
-----------------------------\\\\
a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-----------------------------\\\\
thus\qquad 19^{\frac{7}{4}}\cdot 19^{\frac{b}{a}}=19^{\frac{5}{2}}\cdot 19^{\frac{1}{2}}\implies 19^{\frac{7}{4}+\frac{b}{a}}=19^{\frac{5}{2}+\frac{1}{2}}
\\\\\\
$

$\bf 19^{\frac{7}{4}+\frac{b}{a}}=19^{\frac{6}{2}}\implies 19^{\frac{7}{4}+\frac{b}{a}}=19^3\impliedby 
\begin{array}{llll}
\textit{same base, thus}\\
\textit{exponents must be the same}
\end{array}
\\\\\\
\cfrac{7}{4}+\cfrac{b}{a}=3\implies \cfrac{b}{a}=3-\cfrac{7}{4}$

5 0

2 years ago

Calcuate the distance Tom rides his bike if he rides 5\6 mile of each 5 days

defon

The multi-stage fitness test (MSFT), also known as the beep test, bleep test, PACER (Progressive Aerobic Cardiovascular Endurance Run), Pime, they are cautioned. A second caution ends the test for that runner. The number of shuttles completed successfully is recorded as the score of that runner. The score is recorded in Level.Shuttles format (e.g. 9.5).

8 0

2 years ago

Read 2 more answers

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