Question

The produce manager at a grocery store claims that Granny Smith apples weight a mean of 4 ounces. (FYI: Granny Smith apples are the green ones.) You are a quality control representative and are testing to see if the claim is reasonable or not.

The data below represent a POPULATION of scores from which you are to take a SAMPLE. Your sample size will be ten (10) apples. Your task is to take a sample of 10 scores to test the null hypothesis that the population mean is 4 ounces (Ho: µ = 4) against the alternative hypothesis that the population mean is not 4 ounces (Ha: µ ≠ 4). The "alpha" value (i.e., significance level) will be 0.05. Note that this problem relates to Chapter 11 in our text!

On this assignment you DO NOT HAVE TO SHOW COMPUTATIONS! BUT DO SHOW THE VALUE OF tobt. AND WRITE A ONE SENTENCE CONCLUSION INCLUDING THE P - VALUE.

More Specifically:

1. Write one sentence indicating which apples were chosen chose the data (you must use a random sample using the random numbers generator at the website indicated below).
2. Show the raw data that make up your sample (please put the data in sequence - high to low -thank you!)
3. Write a conclusion -- verbally and statistically (don't forget the p -value!)

The example below is a similar problem I did with a sample of 10 grapefruit, the manager claiming that the mean of the population of grapefruit is 16 ounces.

NOTE THAT I WILL BE ABLE TO TELL IF YOUR CALCULATIONS ARE CORRECT FOR t_obt

Also, be reminded that the real population of apples would be extremely large! There would be thousands -- I only listed 100 numbers below, from which you, again, take 10 (and only 10). Again, use a random numbers generator to choose an unbiased sample!

The random numbers generator can be found at:

https://www.random.org/integer-sets/

for this assignment, you need to generate 1 set with 10 unique random integers. Each integer should contain a value between 1 and 100 and each represent one of the apples of the 100 in the population. I will answer questions about this in class -- feel free to ask!

Here are the data -- these data represent the weights in ounces of the population of Granny Smith apples, each number representing the weight of one apple.

4.3        4.4        4.5      4.8

3.6        4.3        5.0        4.4

3.9        4.2        3.8        4.5

3.9        4.4        3.8      5.2

4.5        4.4        4.4        4.2

4.6        4.2        4,0        4.0

4.2        4.4        4.3        4.2

4.0        4.4        4.7      4.2   

4.0        4.4        4.4        5.4

3.9        4.3        4.2        4.1

4.0        4.4        4.3        4.9

3.7        4.2        4.4        4.4

4.4        4.4      3.8        4.0

4.4        4.2    4.1        5.5

4.4        5.2        4.2        4.6

4.2        3.8        4.0        4.5

4.0        4.7        4.7        4.2

3.8        4.2        4.0        4.0

4.7        3.9        4.0        4.7

4.5        4.4        4.6        4.5

4.3        4.5        3.9        4.3

4.8        4.2        4.2        4.3

3.9        4.5        3.4        4.8

3.9        4.6        4.4        4.4

4.6        4.2        4.0        3.7

This is what a completed, fully-credited assignment looks like (IF IT INVOLVED GRAPEFRUIT -- THE DATA ABOVE DO NOT APPLY TO THIS GRAPEFRUIT EXAMPLE -- I DID NOT SHOW THE GRAPEFRUIT DATA)!!!

1. I numbered each of the grapefruit in the population by calling the upper left value "1" and enumerating from left to right, top to bottom, so the value on the lower right is 100. Then I identified the grapefruit in my sample by using a random numbers generator and chose grapefruit numbered 09, 12, 19, 58, 65, 66, 71, 79, 83, & 99.

2, The raw data

          X

       16.3

       14.9

       15.7

       14.6

       13.9

       16.2

       14.6

       16.3

       15.7

       14.7

3. The population of grapefruit weigh significantly less than 16 ounces
4. (t_{obt (9) =  - 2.62 p <} 0.05)

Answer 1

1.
I numbered each of the apple in the population by calling the upper left value "1" and enumerating from top to bottom,left to right, so the value on the lower right is 100. Then I identified the apple in my sample by using a random numbers generator and chose apple numbered 2, 6, 40, 44, 57, 63, 81, 83, 89, 91.

2.The raw data is
X
3.6
4.6
5.2
3.9
4.3
3.8
4
4.2   
5.5
4.5

The population of apple weigh 4 ounces.
(Z=1.8556, p=0.0635 > 0.05).

Computation in MS-Excel:

Number	Apples weight	Number	Apples weight	Number	Apples weight	Number	Apples weight
1	4.3	26	4.4	51	4.5	76	4.8
2	3.6	27	4.3	52	5	77	4.4
3	3.9	28	4.2	53	3.8	78	4.5
4	3.9	29	4.4	54	3.8	79	5.2
5	4.5	30	4.4	55	4.4	80	4.2
6	4.6	31	4.2	56	4	81	4
7	4.2	32	4.4	57	4.3	82	4.2
8	4	33	4.4	58	4.7	83	4.2
9	4	34	4.4	59	4.4	84	5.4
10	3.9	35	4.3	60	4.2	85	4.1
11	4	36	4.4	61	4.3	86	4.9
12	3.7	37	4.2	62	4.4	87	4.4
13	4.4	38	4.4	63	3.8	88	4
14	4.4	39	4.2	64	4.1	89	5.5
15	4.4	40	5.2	65	4.2	90	4.6
16	4.2	41	3.8	66	4	91	4.5
17	4	42	4.7	67	4.7	92	4.2
18	3.8	43	4.2	68	4	93	4
19	4.7	44	3.9	69	4	94	4.7
20	4.5	45	4.4	70	4.6	95	4.5
21	4.3	46	4.5	71	3.9	96	4.3
22	4.8	47	4.2	72	4.2	97	4.3
23	3.9	48	4.5	73	3.4	98	4.8
24	3.9	49	4.6	74	4.4	99	4.4
25	4.6	50	4.2	75	4	100	3.7

Your sample size will be ten (10) apples. Thus, n=10.

Using random number generator, 1 set with 10 unique random integers, taken from the [1,100] range is:

• Set 1: 2, 6, 40, 44, 57, 63, 81, 83, 89, 91

[The integers were sorted in ascending order.]

A	B	C
	sampled no	sampled Apples weight
1	2	3.6
2	6	4.6
3	40	5.2
4	44	3.9
5	57	4.3
6	63	3.8
7	81	4
8	83	4.2
9	89	5.5
10	91	4.5

		Formula in Excel
AVERAGE	4.377778	AVERAGE(C2:C11)
sample stdev	0.643774	STDEV(C2:C11)
n	10
sigma/root(n)	0.203579
Z	1.855681
value	0.03175	1-NORMSDIST(Z)
p-value	0.063499	2*[1-NORMSDIST(Z)]

The provided sample mean is Xbar = 4.3778
and the sample standard deviation (sigma) is s=0.6438, and the sample size is n = 10.

Null and Alternative Hypotheses:
Ho: μ=4
Ha: μ=not equal to 4
̸This corresponds to a two-tailed test, for which a z-test for one mean will be used.

Rejection Region:
The significance level is α=0.05, and the critical value for a two-tailed test is z_c = 1.96.
The rejection region for this two-tailed test is R = {z: |z| > 1.96}

Test Statistics:
The z-statistic is computed as follows:
z=(xbar-μ)/ (s/sqrt(n)) where, Z~Normal(0,1)
Z=(4.3778 - 4)/(0.6438/sqrt(10)) = 1.8556

Decision about the null hypothesis:
Since it is observed that |z| = 1.8556 < z_c = 1.96, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0635, and since p = 0.0635 ≥ α (0.05), it is concluded that the null hypothesis is not rejected.

Conclusion:
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 4, at the 0.05 significance level.