A p value question from Chapter 9

Consider the following hypothesis test:

Ho:µ=15
Ha:µ not equal to 15

A sample of 50 provided a sample mean of 14.15. The population standard deviation is 3.
a. compute the value of the test statistic.
b.What is the p-value.
c. At a=.05 what is your conclusion?
d. what is the rejection rule using the critical value? what is your conclusion?

Worked solution:

We’re asked to calculate a test statistic. Note carefully that I am going to use the t distribution below----theoretically the normal distribution would work because n > 30, but I am super-careful and always use the t distribution for this.

Recall that we get the test statistic from



So now use Excel: =tdist(2.005,49,2)= 0.0505
degrees of freedom is n-1 = 50 -1 =49
tails is 2 ----look at the hypothesis

The p value is 0.0505. We are asked to test at 95% (meaning alpha = 0.05), so here p NOT smaller than or equal to alpha because 0.0505 > 0.05. Conclusion is fail to reject null hypothesis.

Note: this is very borderline case. A bit of rounding off the other way.....might have been a different conclusion.

The last part of the question asks about using the critical value. In my class I teach you ONLY the p value, so don’t worry about this part of the question.

Norminv and Normdist

Your boss says that 27% of customers order wine that costs above a certain amount. What is that amount. Mean 24, SD4.


Notice the word ‘above’. So we want to find the value of the random variable ‘x’....here the price of the wine....that separates the top 27% of the customers from the bottom---and here’s the key---73%. Recall that Excel goes from left to right. We know the percentage, want to find the value that gives that percentage. In this case, norminv is the boy! So go
=norminv(0.73,24,4)=26.45.

Now, a contrasting problem and more complex. In a restaurant, the bills are normally distributed. The mean bill is 28, and the standard deviation is 6. If 12 of the day’s bills are over 43.06 how many customers did the restaurant have that day?
First, find the percentage of the total bills occupied by the area to the right of 43.06. Here we have the value of the random variable and we want to get the area (or probability, whichever way you want to look at it). Recall that Excel adds from left to right....and our area is clearly to the right. It says ‘above’. So go
=1-normdist(43.06,28,6,true) and get 0.006.


This means that 0.6% of the restaurant’s bills were over 43.06.
Now, we know that 0.6% of the total number of bills is 12 in number. Call the total number of bills ‘L’. We are trying to find L.
0.006*L = 12.
Divide both sides by 0.006 to find L
L = 2000.

Empirical Rule Worked Example

Q6. The vodka dispensing machine keeps malfunctioning. However----it can be regulated to give µ ounces (I don`t know: what measures did they use?). If the ounces of fill are normally distributed with a standard deviation of 0.4 ounces, at what value should we set µ so that a 6 ounce vodka mug will overflow only 2.5% of the time?

Answer: We know from the Empirical Rule that 95% of the observations are within 2 standard deviations of the mean. So that means that 5% of the observations are in the tails to the left and right of 95%. The normal distribution is symmetric, so each of the two tails contains 2.5% of the observations. We want to set the machine so that 6 ounces marks the point where 2.5% of the drinks overflow (dreadful waste of vodka if you ask me!). Look at my beautiful sketch. Make sure you can understand why having 2.5% in one tail means that the random variable that corresponds to that area must be 2 standard deviations from the mean.

We’re given that the standard deviation is 0.4 ounces. Two standard deviations is 0.4 * 2 = 0.8. So µ must be 0.8 ounces from 6 ounces. So we should set µ = 6 – 0.8 = 5.2 ounces.

Number of samples from a population

4. Now your mum wants in on the fun. She collected only eleven candies. How many possible samples of size four could she get from her collection?

Note that N is the notation for size of population and n is the notation for the sample. So in this case, N = 11 and n = 4. The working is below. Use the ! key on your calculator to get the numbers. The symbol ! means ‘factorial’, an instruction to multiply out all the numbers down to 1. So 4! = 4 * 3 * 2 *1 = 24

Chapter 8 Q17

Chapter 8 Q17

Use descriptive stats to get the data below:
Rating

Mean 6.34
Standard Error 0.30587446
Median 6.5
Mode 8
Standard Deviation 2.16285903
Sample Variance 4.67795918
Kurtosis -1.18062982
Skewness -0.14449225
Range 8
Minimum 2
Maximum 10
Sum 317
Count 50
Confidence Level(95.0%) 0.61467772

Now, the question asks us for a 95% confidence interval. n>30 so we can use the CLT. Note that the margin of error given above is for the t distribution. The margin of error for the normal distribution (that is using the CLT) will be a little narrower.
We can calculate this by hand or using Excel. Let’s have fun (!) and do both:
The M of E by hand for 95% is 1.96*0.306 (rounding up) = 0.6
In Excel we go =confidence(=0.05,2.162,50) = 0.6
So the Confidence Interval is the mean plus or minus the Margin of Error:
6.34 – 0.6 to 6.34 + 0.6 or 5.74 to 6.94
Notice that the Margin of Error for the t dist is indeed slightly larger than that for the normal distribution

Chapter 6 Q18

Chapter 6 Q18

Let’s use Excel for this question...although the answers in the book use z scores.

a. At least $40. Recall that Excel sums the probability from left to right, and we want the segment to the right of $40. So go

=1-normdist(40,30,8.2,true) = 0.111

b. No higher than $20. Now we want the segment to the left of $20, so go

=normdist(20,30,8.2,true)=0.111

Now, you may have noticed that the two answers are the same! Well, pretty weird if they weren’t....because this is a normal distribution which is by definition symmetric. The mean is bang in the middle between 20 and 40, so of course the little ‘tails’ either side are the same area.

c. Now we want to know the value of the random variable ‘x’ that creates an area of 10% to the right of it? (Top 10%....) Use norminv for this, but recall that Excel sums from left to right. So go

=norminv(0.9,30,8.2)=40.51. That’s how high the stock price would have to be. See sketch below

Worked hypothesis test example

Here is an example of a typical Stephen P exam question with worked solution

15. You find that you get to love stats so much you want to take more courses! To pay the tuition fees, you get a job in a restaurant as a chef. The boss says that he always budgets for 120gm of smoked salmon in the smoked salmon omelette dish, but you suspect him of lying. You secretly measure the amounts over 42 samples. You find that the mean amount is 117gm with a sample standard deviation of 15gm. This is important: if the boss is lying you could blackmail him and get rich! (8)

a. Write the hypotheses

b. Test at 95%

c. Report your results

d. What are your conclusions?

Solutions....

a.

b. This is a two-tailed test. n > 30 so we could make use of the central limit theorem and use the normal distribution. But for myself I prefer to use the t distribution. So in Excel that’s =tdist(1.296,41,2)

giving you a p value of 0.202

c. We were asked to test at 95%, giving an alpha of 0.05. The p value that we calculated is larger than alpha (0.202 > 0.05) so following the rejection rule we fail to reject Ho.

d. Unfortunately (!) the boss is not lying so can’t blackmail him!

Chapter 8 Q1 Q2

Chapter 8 Q1

We are told the population standard deviation, so we can use normdist...or calculate by hand!

Chapter 8 Q2

The little graphics below shows how to calculate the 95% interval by hand. In Excel first find the Margin of Error:

a. =confidence(0.1, 6,50) = 1.396

b. =confidence (0.05,6,50) = 1.67

c. = confidence (0.01,6,50) = 2.19

THEN find the Confidence Interval by adding and subtracting from the mean (which is given at 32)...don't forget this part!

Chapter 7 Q19

Chapter 7 Q19

a. The population standard deviation for this problem is 4000 (see page 278)

We can use Excel to find first the probability that the sample is in the area UP TO 51800 + 500 = 52300 and then UP TO 51800 – 500 = 51300. First find the standard error for a sample size of 60. This is 516.4. Then use normdist

=normdist(52300,51800,516.4,true) = 0.83

=normdist(51300,51800,516.4,true) = 0.17

then subtract to get 0.66

b. For (b), recalculate the standard error with the sample size of 120.

SE Notation

The symbol for a population standard deviation is the Greek letter σ (‘sigma’).
For a sample standard deviation the symbol is s.

The notation for a standard error is given on the right of this post.

We don’t change the symbol on the left. Just switch the sigma in the numerator on the right side to s if you happen to know that it is from a sample (for example you just calculated it from data that is a sample)

Chapter 8.26

Chapter 8.26

Recall that the Margin of Error for 95% is
ME = 1.96*SE. Now, make sure you can follow the math here:

Chapter 8.22

Chapter 8.22

Note we’re told that the distribution has a normal distribution. This is important because the sample size is only 25, less than 30. Use Excel to get the mean

a. 3.348
b. Use =confidence(0.05,2.287,25) to get the margin of error. This is 0.896 (rounded). So the Confidence Interval is 3.348 – 0.896 to 3.348 + 0.896 (finish it off yourself which in the formal answer in a test you MUST do).

Now, the question doesn’t specify whether you need to calculate the Margin of Error ‘by hand’. Let’s do that just for practice. Recall that the M of E for 95% is 1.96 * SE. Here the Standard Error --- SE --- is 2.287/5 = 0.457 (rounded). So the Margin of Error is 1.96*0.457 = 0.896. The same as when we used =confidence. Then you can get the Confidence Interval in the same way....

Chapter 8.13

Chapter 8.13

Note that the sample size is only 8 and you don’t know whether the population from which the sample is drawn is normally distributed or not. In this situation use the t distribution. Put the numbers into Excel and then find the margin of error using the t distribution.....go to data analysis> descriptive stats and it is the number at the bottom of the table.

My results are:

Column1

Mean 10
Standard Error 1.224745
Median 10.5
Mode #N/A
Standard Deviation 3.464102
Sample Variance 12
Kurtosis -1.04286
Skewness -0.16496
Range 10
Minimum 5
Maximum 15
Sum 80
Count 8
Confidence Level(95.0%) 2.896061

So for 13 (a) xbar is 10; for (b) 3.464, and (c) is 2.896. The last question is the Confidence Interval. That’s easy: 10 – 2.896 and 10 + 2.896....note you MUST actually finish off this little calculation.

Difference between cluster sampling and stratified: think of cluster being used for spatial problems, such as the example about trees with beetles in BC. Stratified is used when we can easily divide the population into homogeneous groups, for example golf-club membership