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3 years ago

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Assume you have a sample of n1=8, with a sample mean X1= 42 and a sample standard deviation, S1 = 4 and an independent sample of

n2=15 from another population with a sample mean of X2=34 and a sample standard deviation of S2 = 34. The pooled variance, Sp 2 = 776.
a. What is the value of the tSTAT for testing H0 : µ1 = µ2?
b. In finding the critical value, how many degrees of freedom are there?
c. What is your statistical decision?

1 answer:

patriot [66]3 years ago

5 0

Answer:

Step-by-step explanation:

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It was recently estimated that females outnumber males by about three or two. If there are 1770 people in a county, how many of

max2010maxim [7]

GIVEN

1) Females outnumber males by 3 to 2.

2) There are 1770 people in the city.

SOLUTION

Let the number of males be x and the number of females y. Therefore, the total number of people can be written to be:

$x+y=1770\text{ -----------(1)}$

If there are 3 females for every 2 males, we have that:

$\begin{gathered} \frac{y}{3}=\frac{x}{2} \\ \therefore \\ x=\frac{2y}{3}\text{ ----------(2)} \end{gathered}$

We can substitute equation (2) into equation (1) and solve for y:

$\begin{gathered} \frac{2}{3}y+y=1770 \\ \frac{2y+3y}{3}=1770 \\ 5y=1770\times3 \\ 5y=5310 \end{gathered}$

Dividing both sides by 5, we have

$y=1062$

There are 1062 females in the county.

3 0

1 year ago

One kilogram is approximately equal to 2.21 pounds. Find the number of pounds in 165 kilograms. Round to the nearest tenth of a

Shtirlitz [24]

Answer: 364.7 pounds

Step-by-step explanation: Multiply the number of kilograms, 165, by the number of pounds per kilogram, 2.21.

165 x 2.21 = 364.65

Round to the nearest tenth.

364.7 pounds in 165 kilograms.

8 0

3 years ago

A football team loses 5 yards on each of 3 consecutive plays. What signed number represents the team's total loss of yards on th

mario62 [17]

Answer:

change below :)

Step-by-step explanation:

You would start out at in the middle of a number line (horizontal). After the first play, you would move to the left on the line 5 spots (to simulate 5 yards lost) and then make a dot to show your new position. You would do the same thing for the next two plays, moving 5 spots to the left each play and creating a new starting point each time. The total spots moved after the 3 plays is 15 spots to the left.

3 0

2 years ago

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determinin

riadik2000 [5.3K]

Answer:

Option b - not significantly greater than 75%.

Step-by-step explanation:

A random sample of 100 people was taken i.e. n=100

Eighty of the people in the sample favored Candidate i.e. x=80

We have used single sample proportion test,

$p=\frac{x}{n}$

$p=\frac{80}{100}$

$p=0.8$

Now we define hypothesis,

Null hypothesis $H_0$ : candidate A is significantly greater than 75%.

Alternative hypothesis $H_1$ : candidate A is not significantly greater than 75%.

Level of significance $\alpha=0.05$

Applying test statistic Z -proportion,

$Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

Where, $\widehat{p}=80\%=0.80$ and $p=75%=0.75$

Substitute the values,

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}$

$Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}$

$Z=\frac{0.05}{0.0433}$

$Z=1.1547$

The p-value is

$P(Z>1.1547)=1-P(Z$

$P(Z>1.1547)=1-0.8789$

$P(Z>1.1547)=0.1241$

Now, the p-value is greater than the 0.05.

So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.

Therefore, Option b is correct.

7 0

3 years ago

Can someone help me please?

Rashid [163]

Idk your best bet is to look it up online thats what i do often

8 0

3 years ago