Can I get some help with questions? (1 Viewer)

jimmysmith560

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In general, we reject the null hypothesis if the p-value is less than the significance level α.

If the alternative hypothesis contains the not‐equal‐to symbol (≠), the hypothesis test is a two‐tailed test. Otherwise, all other tests are one-tailed.

If needed, below are solutions to each part, based on the above information. As cossine mentioned, definitely try to attempt the question(s) first to see if you can apply your understanding in your working:

Part (a):

Suppose is the average age of Sydney residents; then is , and is . Note that this is a one-tailed hypothesis test. Here, the p-value 0.0170 is less than the significance level . So we reject the null hypothesis.

Conclusion: this sample information does indicate that the mean age in Sydney has increased from 35 years.

Part (b):

Let be the average weight of this population; then is , and is . This is a one-tailed test. The standard deviation of weights in this population is . Here, the p-value 0.001 is less than the significance level . So we reject the null hypothesis.

Conclusion: the sample data do provide enough evidence for us to conclude that the mean weight for the population is less than 70 kg.

Part (c):

Let be the average IQ score of this population; then is , and is . This is a two-tailed test. Here, the p-value 0.095 is greater than the significance level . So we do not reject the null hypothesis.

Conclusion: on the basis of these data, we cannot conclude that the mean IQ score for this population is not 100.

I hope this helps! :D
 

cossine

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Can it also be the greater than or equal to symbol?
3:37 of this video
So if you have

H0: mean_weight >= 70
Ha: mean_weight <70

Reject the null hypothesis would be equivalent to rejecting H0: mean_weight=70

For computing the z-value you would have mean_weight=70.
 

jimmysmith560

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View attachment 35554

I solved Q.6 and got 21. If you count the 2 ones in 11.

But I don’t know how to do Q.7
Your answer to the first question is correct.

I was not able to find the official solution to the second question unfortunately. However, I found that there are 537 1's that occur in the numbers 1 to 1150. If the question is asking for the total number of 1's between 1 and 1150 (inclusive), then 537 would be the answer. If the question means how many times you will see the number 1 printed (i.e. same as the previous question), then you will need to find that number, which will be lower than 537. Perhaps someone else would be able to help you with this question more closely.

I hope this helps! :D
 

jimmysmith560

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Is there any way to find how many day each man works and solve it like that?
Essentially, the higher the number of men, the lower the time it will take to dig a pond. Similarly, the lower the number of men, the higher the time it will take to dig a pond.

There is some form of proportion/negative relationship between the number of men and the number of days. For example, having half the number of men means that it will take twice the time to dig a pond, i.e. . Of course, there can't be 7.5 men in reality but this is correct from a mathematical perspective. Another example is if you have 20% the number of men (20% of 15 is 3), in which case the time it will take to dig a pond will be multiplied by 5, resulting in . Consequently, if the capacity becomes 15 times lower, then it will logically take 15 times the amount of time to dig a pond, leading us to finding that , i.e. it will take 1 man 300 days to dig a pond.

Now that you have this information, you can solve the question algebraically, as follows:





.

 

cossine

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Why is that? Or is just the way the math works
I had another look at the question. I was not unfortunately able to understand the other answers.

Theorem: if A and B are independent then P(B|A') = P(B) # This is actually converse of what we are looking to prove. Anyhow.

Proof:

P(A)*P(B|A) + P(A')*P(B|A') = P(B)

Let P(A) = p, let P(B) = q


=> P(A')*P(B|A') = q - pq

=> (1-p)*P(B|A') = q(1-p)

=> P(B|A') = q

This suggests the answer is A.
 

jimmysmith560

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the angle of elevation from point a is 14 degrees, the angle of elevation from point b is 20 degrees. the length between point a and b is 100m, find the height.

i tried using substitution like my tutor told me to but i keep getting a negative answer??

pls help

3D515078-3159-4A69-8F7E-777397D81D4B.jpeg
I think that there might be some additional information that would be useful to include so that other people can assist you, such as:
  • Whether the diagram is meant to be a triangle.
  • Your working/attempt at solving this question, which would allow others to determine any errors that you might have made.
Also, correct me if I am wrong, but if the angle of elevation from point A is 14°, then, given the straight line, the angle on the other side of point A must be 166°. However, if the angle of elevation from point B is 20°, then something does not seem right because 166 + 20 = 186°, exceeding the sum of angles of a triangle (with reference to the smaller triangle to the right). I am not entirely sure as to the relevance of this in terms of solving the question, but I thought I should point this out.
 

jimmysmith560

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Hivaclibtibcharkwa said:
1706429534607.png

What would the null hypothesis be for this?
The null hypothesis is "The road users are not inclined to speed through traffic lights if they change from green to amber" i.e, the average time taken to cover last 50 m is 3.6 s.

Consequently, the alternative hypothesis is "The road users are inclined to speed through traffic lights if they change from green to amber" i.e, the average time taken to cover last 50 m is not 3.6 s

I hope this helps! :D
 

jimmysmith560

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kkk579 said:
didn’t have solutions so was wondering if anyone could potentially take a look at my answer and see whether it’s right or not. thank you!!!

24EB6F4C-369A-404F-9496-E68B05D6F207.jpeg

image.jpg
I just had a look at separate working for each part of the question and compared it to your working and answers. Fortunately, all your answers are correct.

I hope this helps! :D
 

jimmysmith560

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kkk579 said:
An aeroplane flies 138 km in a southerly direction from a military air base to a drop-off point. The drop-off point is 83 km west of the air base. Find the bearing, correct to the nearest degree, of:

a. the drop-off point from the air base
b. the air base from the drop off point
In terms of the first question, it appears that you can get the textbook answer by drawing the triangle differently and using inverse cos, as follows:

1657708884468.png



The third angle is the bearing from the drop to the base, so that is

The bearing from the base to the drop is due south (180) plus the other part of the 90 going out west, meaning that

I think that the tricky bit is to note that the question says "in a southerly direction" as opposed to "due south".

I hope this helps! 😄
 

jimmysmith560

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inverse cos? not sure what that is since I’m in year9 and I’ve only been taught right angle trig
Sorry I forgot that you are in year 9. Essentially, inverse cos just means . It is used to measure the unknown angle when the length of two sides of the right triangle are known, as shown above.

Perhaps someone else could explain it better :)
 

jimmysmith560

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kkk579 said:
Regular NSW number plates come as black on yellow or black on white, both with 4 letters and 2 digits.

What is the probability of a plate having 4 letters the same and 2 digits the same?
Would the following working help with the question in the first attachment?









 

cossine

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yes tysm do u mind solving the second one too pls?

Two dice are thrown onto a table and the number on the upper most faces examined. What is the probability that one number will be a factor of the other number?

A. 1/5
B. 1/9
C. 11/18
D. 1/6
Draw a table or just consider the various cases.

You just need to give it a go.

e.g.

1 is a factor of 1
1 is a factor of 2
etc

Write all the options with a denominator of 36. Therefore A) is eliminated as it is not possible to have a decimal number of outcomes. B,D are eliminated because there are more than 6 outcomes of when event the occurs.

Therefore C.
 

jimmysmith560

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Original Poster said:
how do i find the red marked part? i tried labelling everything but i cant seem to able to find that specific angle. i have uploaded my diagram and the q does someone mind helping me??
thanks

image.jpg image.jpg
I would suggest having a look at the following working. The diagram appears to be drawn slightly differently from yours.

1662187447465.png

Very similar versions of this question also appeared in a Baulkham Hills High School trial paper as well as the Cambridge textbook, where the diagrams provided look like this:

BHHS paper:
1662187751637.png

Cambridge textbook:
1662187709391.png

I hope this helps! 😄
 

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jimmysmith560

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Original Poster said:
thanks alot this makes much more sense my diagram was wrong no wonder it was so confusing
Original Poster said:
do u mind telling me what bhhs paper it was from or sending it to me pls
it would be greatly appreciated
No worries!

It is included in the 2009 BHHS 2U trial paper, although BHHS's version of this question does not include part iv (the Cambridge version includes all parts). Either way, you can access it using the following link:

 
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5uckerberg

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So give the question a go first and I will check the working out.

In this case

H0: mean_weight = 70
Ha: mean_weight <= 70.
I believe for accuracy you should put a tilde above because is the boundary condition for the question. Like this
 

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